A girl throws a rock horizontally from the top of a cliff 98 m high, with a horizontal velocity of 27 m/s.?
815192
d = 1/2 a * t^2\n98m =(1/2) * 9.8m/s/s * t^2\n20 sec^2 = t^2\nt = 4.47 sec\n\nd = v * t\nd = 27m/s * 3.16sec\nd = 120.7 m
888567
a) zero momentum, as velocity on top is zero. p=mv\n     p=0.100kg * Om/s\n\nb) Vf = Vi + at\n      0=15+(-10)t\n     t=1.5ec (to reach maximum height)\n\nDivide by 2 for halfway up .i.e 0.75ec\n\nVf = Vi + at\nO=15 + (-10)(0.75)\nVf = 7.5m/s\n\np=mv   (p = momemtum)\n\np = (0.100kg)(7.5m/s)\np = 0.75kg*m/s, upwards (Direction is important)
680534
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
782512
You need to use kinematics and Newton's 2nd law to solve the problem.  In the horizontal direction, Newton's 2nd law states:\n\nForce - friction = m*a.  Since you are given the values, you can solve for the accleration:\n\n1149 - 909 = (2700)*a, solving for a, you get:\n\na = 0.089 m/s^2\n\n----------------\n\nNow use kinematics to solve for the distance needed to reach the desired final velocity.  Use the following kinematic equation:\n\n2ax = v_f^2 - v_i^2.  You know that:\n\nv_f = 3.2 m/s\nv_i = 0 m/s (starting from rest)\na = 0.089 m/s^2\n\nNiw solve for x:\n\nx = [v_f^2 - v_i^2]/(2a) = 3.2^2/(2*0.089) = 57.6 meters\n\n---------------\n\nHope this helps
735201
He would need to throw at least 129 MPH to do that. Here is my calculation:\n\nAssume he throws overhand so that he releases the ball near the ceiling, 8 feet above the floor. Assuming he throws parallel to the floor, the ball starts 8 feet up and begins accelerating down at 32 feet per second square as soon as he releases it. Using the formula X=1/2AT^2, I calculate it takes 0.707 seconds to hit the floor, so it has to travel 133.5 feet horizontally in less than 0.707 seconds to hit the far wall before hitting the floor. 133.5 feet in 0.707 seconds is 189 feet per second or 129 MPH.\n\nBut wait, there is more!\n\nIf he throws underhand, releasing the ball near the floor, and if he throws at a slight up angle so the the ball nearly touches the ceiling half way down the hall, then it takes twice as long to hit the floor. It takes 0.707 seconds to rise up to 8 feet and another 0.707 seconds to fall back to the floor. So he would only have to throw half as fast to hit the far wall; only 64 /12 MPH.\n\nSo the question becomes, can he throw 129 MPH overhand or 64 1/2 MPH underhand?
43647
Let the mass of the girl be m\nLet her net volume be v\ndensity in air = d\n\nSince 97% of her body is beneath water, so volume equal to 97m/100 is displaced and weight of 97m/100 mass is balanced by buoyant force. \n\nSo, here mass density remains the same = d = m/v\n\nBut her weight density = (mg- [97mg/100])/v\n                       =  3mg/100v\n                       =  3% of weight density in air
1129113
It is 9.8 m/sec , some they have made it 10 :)
973163
Relative to a woman in the car:\n\n(a) the translational velocity is ZERO since the woman is moving at the same translational velocity as the car relative to the ground.\n\n(b) the magnitude of the translational acceleration of the center of the wheel is also ZERO because the center of the wheel is NOT moving and NOT rotating relative to the woman.\n\n(c) v, velocity of the point at the top of the tire is = 80 km/h because that is how fast the car is moving, thus the wheel must also be rotating at the same speed.\n\n(d) a, acceleration of the point at the top of the tire is given by:\n\na = v^2/r = (22.22)^2/(0.33) = 1,496.45 m/s^2\n\n(e) v, velocity of the point at the bottom of the tire is = -80 km/h because at the bottom of the tire, the tire is moving in the opposite direction\n\n(f) a, acceleration of the point at the bottom of the tire is the same as (d) a = v^2/r = 1,496.45 m/s^2\n\nNow relative to a hitch hiker on the ground:\n\n(g) v at the center of the wheel is = 80 km/h or 22.22 m/s, since the car and the center of the wheel (which is not moving relative to the car) are moving at the same speed relative to the ground.\n\n(h) a at the center of the wheel is = v^2/r = 1,496.45 m/s^2, since the center of the wheel is rotating about the bottom point of the tire as the tire rolls on the ground.\n\n(i) v at the top of the tire is v(top) = ωd, where ω is the angular velocity of the rotating tire and d is the diameter of the wheel. The reason we use diameter of the wheel is because the top of the tire is rotating abou the bottom of the tire, not the center.\n\nω = v/r = (22.22)/(0.33) = 67.33 rad/sec\nv(top) = ωd = (67.33)(0.66) = 44.44 m/s\n\n(j) a at the top of the tire is = v(top)^2/d\n    = (44.44)^2/(0.66) = 2,992.29 m/s^2  \n\n(k) v at the bottom of the tire is equal to ZERO because the bottom of the tire is not moving relativ to the hitch hiker or the ground.\n\n(l) a at the bottom of the tire is also equal to ZERO since the velocity is zero.
812983
I solved this as if the artist strikes the trampoline at 5 m/s downward.\nIs he standing on the trampoline and gets sent upward at 5 m/s, or is he standing on a platform some distance above the trampoline?\n\nFor striking the trampoline at 5m/s downward\nStart by considering the energy in the system:\nThe artist has kinetic energy of 1/2*m*v^2\nand potential energy of m*g*d\nwhere g is 9.8 m/s^2 and d is the net downward displacement of the spring\n\nThe spring will absorb the energy =1/2*k*d^2\nwhere k is the spring constant\nat equilibrium when the spring is fully compressed the energy in the spring will balance out the rest\nset all energy equal to zero for equilibrium\n0=1/2*kd^2-m*g*d-1/2*m*v^2\nmultiply by two\n0=k*d^2-2*m*g*d-m*v^2\n\nplug in the numbers\n0=6.2E+4*d^2-2*65*9.8*d-65*5^2\n0=6.2E+4*d^2-1274d-1625\nthe positive root, which is the displacement of the spring, \nset negative since I used positive motion downward\nd=-.172m\n\nSince this isn't the answer, I will assume that the artist was on a platform and jumped onto the trampoline.  I will compute the height of the platform above the uncompressed trampoline:\nIn this case the spring energy is computed from your answer:\n1/2(62000)*.31^2\n=2979\nthe potential energy lost due to compression is\n65*9.8*.31\n=197.5\nHis vertical velocity when he jumped from the platform\n=.5*65*25\n=812.5\nThe energy will sum to zero.\nThe energy in the spring came from three sources:\nHis vertical velocity in kinetic energy\nPotential energy by standing on a platform above the trampoline\nThe potential energy given to the spring due to his weight and travel during compression\nSo\n2979-197.5-812.5-65*9.8*h=0\nwhere h is the height of the platform\nh=1969/637\nh=3.1m\n\nj
290866
That is the realm of theoretical physics my friend with no proven answers, only theory.  As for inquisitive1 your thought question doesn't adress the full idea.  if an egg was brought to a high enough velocity, it could crush a rock.
1033468
2.36 m s^-1 to 2 d p\n\nWe split the launch velocity of the ball into a vertical and a horizontal component.\nHorizontal component = 19.8 * cos(49.1) = 12.9639 m/s\nVertical component = 19.8 * sin(49.1) = 14.9659 m/s\n\nGravity accelerates the ball vertically with the acceleration -9.8 m s^-2 (negative because the pull of gravity is downwards while the ball is moving upwards to start with)\n\nThe vertical distance from the ball to the ground after time t is given by the formula s = u * t + (g * t^2) / 2 where: u = initial velocity, g = acceleration of gravity, t = time elapsed\n\nWhen the ball returns to the ground, s = 0\nTherefore the time taken for the ball to return to the ground is given by\nu * t + (g * t^2)/2 = 0 Substitute values\n14.9659 * t - (9.8 / 2) * t^2 = 0 Remove brackets, divide by t, rearrange\n4.9 * t = 14.9659\nt = 3.05427 s\n\nDuring that time the ball will travel horizontally \ns = u * t where s is distance in m, u is velocity in m s^-1, t is time in sec\nDistance travelled horizontally is 12.9639 * 3.05427 = 39.5953 m\n\nThe player has to run 46.8 - 39.5953 m = 7.20475 m in 3.05427 s\nHis speed must therefore be 2.35891 m s^-1
528749
Humans don't generally bounce, so I'm thinking that if you drop her off the top of a skyscraper, she's going to go splat like a tomato.  Maybe she'll bounce around 10-30 centimeters due to the "recoil" effect of her hitting the ground.  Don't kill your boss; just quit your job if you are thinking about it.
281360
its a rock...\n\npencils...
905071
The trick to solving this problem is to set up your equations properly:\n\nR = Vx * t\n\nH = Vy * t - (1/2)g * t²\n\nWhere R is the projectile's range and H is its initial height, Vy is the projectile's initial vertical velocity and Vx is its horizontal velocity.\n\n[[sorry for the edit, I just noticed the 45° angle; that changes the analysis]]\n\nHere's what we'll do: let's flip the problem around and pretend that instead of landing at a downward angle of 45°, it takes off at an upward angle of 45° with some velocity V, and at some point, t, the projectile has the coordinates (50, 40).\n\nThis introduces two new equations:\n\nVx = V * cos(theta)\nVy = V * sin(theta)\n\nCombining the equations now provides:\n\nR = V * cos(theta) * t\nH = V * sin(theta) * t - (1/2)g * t²\n\nSolving for t in equation 1:\n\nt = R / (V * cos(theta))\n\nSubstituting this into equation 2:\n\nH = V * sin(theta) * R / (V * cos(theta)) - (1/2)g * (R / (V * cos(theta)))²\n\nH = R * tan(theta) - (1/2)g * (R / V)² * (1/ cos(theta))²\n\nWe can now solve for V as:\n\n-H * cos(theta)² + R * tan(theta) =  (1/2)g * (R / V)²\n\n[-H * cos(theta)² + R * tan(theta) * cos(theta)²] =  (1/2)g * (R / V)²\n\nV² = (1/2)g * R² / [-H * cos(theta)² + R * tan(theta) * cos(theta)²]\n\nV = [(1/2)g * R² / [-H * cos(theta)² + R * tan(theta) * cos(theta)²]]^(1/2)\n\nSubstituting the values provides:\n\nV = 49.497 m/s\n\nFinally, plugging this back into the expression for Vx:\n\nVx = V * cos (theta)\n\nVx = 49.497 * cos (45°)\n\nVx = 35 m/s\n\nIf we now, flip our coordinate system back around, we know that the horizontal component of velocity is 35 m/s (which does not change over time). However, this implies that there will be some vertical component of velocity when the projectile is fired. We find this by knowing that the final vertical component Vy equals the final horizontal component (to form the 45° angle):\n\nVx = Vy\n\nAdditionally, the vertical component of velocity changes due to gravity. Thus, the initial vertical velocity component is:\n\nVyo = Vy - g*t = Vx - g*t\n\nWhere t = R/Vx = 50/35 = 1.42 s. Plugging this in, we get:\n\nVyo = 21 m/s (downward)\n\nFINAL RESULTS:\n\nVx = 35m/s\n(Horizontal Velocity)\n\nVy(t) = -21 - 9.8*t\n(Vertical Velocity as a function of time)\n\nH(t) = 40 - 21*t-(1/2) * g *t²  \n(Projectile Height as a function of Time)\n\nR(t) = 35 * t\n(Range as a function of time)\nHope that clears things up a bit! :)\n\nVo = sqrt(Vx² + Vyo²) = 40.82 m/s\n(Launch Velocity)\n\nLaunch Angle = arctan(Vyo/Vx) = -30.96°\n\nTime of Flight = 50/35 = 1.42 seconds\n\n-----------\n\nYes, it's long, I know. But I hope it clears stuff up for you :)
643339
It's all about conservation of energy.   The contestant did some work on the ball.   Remember the definition of work?\n\nWork = Force * Distance\n\nSo, the contestant did a work:  \n              W = 810N * 2.0m = 1620Joule\n\nOk, since energy is conserved, the ball will have a kinetic energy of 1620Joule at the begining of the throw, but it will eventually lose it as it goes up, transformed into Potential energy.\n\nWhen the ball reaches the highest point its speed is equal to zero, and hence the kinetic energy is zero too.  Conservation of energy requires that all 1620Joules transformed into potential energy.  So, do you remember the formula for potential energy?\n\nP. energy =  mass * gravity * height ; \n\nwith gravity = 9.82m/sec^2\n\nSince the potential energy is equal to 1620Joules:\n\n1620Joules = (7.0Kg)*(9.82m/sec^2)*height\n\nSo:\n\nheight = (1620Joules)/(7.0Kg * 9.8m/sec^2) = 23.567m\n\nHope I helped
1119837
First divide up the initial velocity into its horizontal and vertical components.\nV is the initial velocity of 40 m/s.\nTheta is the launch angle of 37 degrees above the horizon.\nV_x_i and V_y_i are the horizontal and vertical components of the initial velocity respectively.\n\nV_x_i = V * cos(theta) = 40 * cos (37) = 31.95 m/s\nV_y_i = V * sin(theta) = 40 * sin (37) = 24.07 m/s\n\nAlthough the vertical component of the object's velocity will change as the object experiences gravitational acceleration toward the ground, neglecting air resistance, the horizontal component of velocity can be treated as a constant.\n\nV_x = V_x_i = 31.95 m/s\nV_y = V_y_i  - g*t\nwhere t is the time after the object is launched and g is the gravitational acceleration experienced by the object (g = 9.81 m/s^2).\n\nIf the fence which the object hits is 128 meters away (in the +X direction), then the object (traveling at 31.95 m/s in the +X direction) will hit it at about 4.01 seconds after launch.\n\nBy taking this value (4.01 s) and plugging it in for t in the below shown equation to solve for the distance relative to the initial launch height the object has traveled,\nd = V_y_i * t + 1/2 (g)t^2\none can calculate that the object is at a height of about 17.65 meters above its original launch height.\n\nSo I calculate the impact height to be 17.65 meters.
698592
The acceleration due to gravity is independent of the mass of the object.\n\n The speed of any mass thrown up will decreases by an amount 9.8 m/s per second. \n\nIf the initial speed is u, then its kinetic energy is ½ m u u. Its potential energy is zero.\n\nThe total energy is½ m u u + zero = ½ m u u.\n\nAs it goes up its kinetic energy decreases whereas its potential energy increases keeping the total energy same.\n\nWhen it goes up through ‘h’ meter,  its total energy is [ ½ m u u + mgh].\n\n[ ½ m u u + mgh] = ½ m v v.\n\nm u u + 2mgh = m v v. \n\nCancelling m on both sides,\n\nu u + 2gh = v v\n\nThus we have the familiar equation vv = uu + 2gh.\n\nWhen the ball reaches 9 m its speed becomes zero.\n\nHence,  v v = 2 x 9.8 x 9\n\nSolving we get v = 13. 3 m/s . \n\nThe speed with which it is thrown up is 13. 3 m/s.\n\nWhen the height is 8 m, now having known the initial speed we use the same equation for solving for uu.\nvv = uu + 2gh\n\n13.3 x 13.3 = uu + 2 x 9.8 x 8 .\n\nSolving for u, we get  u u = 196 or u = 4.43 m/s.\n\nThe problem can be viewed in other way.\n\nWhen the ball is at its heighest point namely at 9m its potential energy is mgh. where h is 9m\n\nWhen it falls through one meter its kinetic enery increases and potetial energy decreases , keeping the total energy as the same.\n\nmg x 9 = ½  m vv + mg x 8.\n\nOr mg ( 9-8) = ½  mvv \n\ng x 1 = v  v/2\n\n2 g = v v. \n\nv v = 2 x 9.8 = 19.6 (m/s)^2. \n\nv = 4.43 m/s.\n\n Note that the mass gets cancelled and hence we need not know the mass of the ball.