The height h in feet of an object after t seconds is given by the function \n   h = -16t2 + 70t + 3. \nHow long will it take the object to hit the ground?  \n\nHow would i solve this?
213368
When it hits the ground h = 0 \nHence the equation reduces to\n16t^2 - 70t -3 = 0\nt = (70 plus or minus sqrt(70^2 - 4*16*(-3)))/2*16\n  = (70 plus or minus sqrt(5092))/32\n  = (70 plus or minus 71.35)/32\nt= 4.42 s approx  (note that 70 - 71.35 is not applicable since it is negative)\nAnswer 4.42s
43880
I'm just doing all your work aren't I?  Its ok I like physics, and its good practice.  I am studying for the MCAT coming up this April (Med School).  This problem unlike the other ones that I did is a little more tricky to me.. Mostly because I don't have my physics textbooks with me.  But it would appear that you need to calculate the acceleration that is exhibited on the foot when it hits the ground.  The way I approached this problem is by using vf^2 = vi^2 + 2ad  and I solved for acceleration.  Knowing that the final velocity is zero I isolated acceleration to get and acceleration of -1024 m/s^2!!! thats huge isn't it.  that is why I'm uncertain about this problem... anyways, multiplying that using the F =ma equation I got a total force of 9216N.  But after inspecting the inital velocity its got to be right because if you convert 6.4 m/s into km/hr you get 23.04 km/hr!!! This person must be falling of a building or something!  So there it is.  Afterall didn't the question start out saying that injury can happen!?
11962
The answer is A. You mean to tell me you couldn't figure this out for yourself?\n\nHere's how it's done.\n\n1. The unit of measure is feet per hour (ft/h).\n\n2. There are 60 minutes in an hour.\n\n3. 30 minutes x 2 = 60 minutes, which is an hour, as mentioned in #2 above.\n\n4. Since you multiplied 30 minutes by 2, you should likewise multiply 1000 feet by 2. 1000 feet x 2 = 2000 feet. There's your answer. Got it?
502317
Basically, in conservation of energy, you have:\n\nEnergy(initial) = Energy(final).  This becomes\n\nKE(initial) + PE(initial) = KE(final) + PE(final), neglecting air resistance.  When the projectile hits the ground, there is no potential energy because h = 0.  Therefore, the equation simplifies to:\n\nKE(i) + PE(i) = KE(f)\n(1/2)m*v_i^2 + m*g*h = (1/2)m*v_f^2.\n\nWe know that:\n\ng = 9.8 m/s^2\nv_i = 175*sin(45) m/s (because we are only taking the vertical component)\nh = 165 m\n\nThe masses cancel, now solve for v_f:\n\nv_f = sqrt([v_i*sin(45)]^2 + 2gh) = sqrt(175^2 + 2*9.8*165)\n\nSolving, you get v_f = 184 m/s\n\n------------\n\nHope this helps
49670
The struggle here is that you provide too little information to recognize this problem as coherent.\n\nWe may assume that the acceleration of gravity is 9.8 m/s/s or 32 ft/sec/sec.\n\nThe velocity(V) of an object is equal to its initial velocity(v) plus its acceleration(a) times time(t). \nV = v + at or V = v - 32t (in feet per second)\n\nYour object will reach maximum height when the velocity is zero or when the initial velocity is equal to 32 ft/sec/sec * time. With your list of times, you'd just multiply by 32 to solve for the initial velocity, if that is what the problem is actually looking for. There are many other permutations of this data, but hopefully that is a start for you.\n\nGood luck.
364210
initial vertical velocity = 30 sin 36.0 degrees = 17.6 m/s\n\nv = u + at\n0 = 17.6 + (-9.81)t\nt = 1.794 s\n\ntime that the ball is in the air = 1.794*2 = 3.588s (since it takes the same time for the ball to rise up to max height and for the ball to fall from the ground to max height )\n\n-----------------------------------------------------------------------------------\ns = ut + 1/2at^2\ns = (17.6)(1.794) + (0.5)(-9.81)(1.794^2)\ns = 15.85 m\n\nthe ball rises to 15.85 m\n-----------------------------------------------------------------------------------\n\nhorizontal speed = 30 cos 36.0 = 24.27 m/s\n\nhorizontal distance = 24.27(3.588) = 87.08 m\n\n------------------------------------------------------------------------------------\n\nplease do redo the calculations yourself cuz i didn't take much care of significant figures and i may have made errors when pressing the calculator
424187
a(t) = -32\n\nv(t) = ∫a(t) dt\n= -32t + c\n\nWhen t = 0, v = 96\nso v(t) = 96 - 32t\n\nX(t) = ∫v(t) dt\n= ∫(96 - 32t) dt\n= 96t - 16t² + k\n\nWhen t = 0, x = 0 so k = 0\n\nSo X = 96t - 16t²\nX max is at the vertex of this parabola\nWhen t = -b/2a \n= -96/(2 * -16)\n= 3 sec\n\nSo Xmax = 96*3 - 16*3²\n= 144 ft
494569
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 :)
636091
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.
482440
As is usual for these sorts of problems, we'll neglect air resistance.\n\nFor part (b) note that at the top of its trajectory, the velocity of the ball is zero, and the kinetic energy it had when it was first thrown has been convered to gravitational potential energy:\n\n0.5*m*(29.4 m/s)^2 = m*g*h\n\nwhere m is the mass of the ball\ng is the acceleration of gravity (assumed constant over the distance the ball travels)\nh is the maximum height\n\nSimplifying:\n\n432.18 (m/s)^2 = g*h\n432.18 (m/s)^2 / 9.81 m/s^2 = h\n44.07 m = h\n\nFor part a, write the position of the ball as a function of time, (assuming that the ground surface is at y = 0):\n\ny(t) = v0*t + 0.5*a*t^2\ny(t) = 29.4 m/s * t - 0.5 * 9.81 m/s^2 * t^2\ny(t) = t*(29.4 m/s - 4.9 m/s^2 * t)\n\nWe want to know the times when y(t) =0 (i.e., when the ball is at ground level).  Obviously, one of these times is t = 0.  \n\nThe other zero of this equation is found from:\n\ny(t) = 0 = (29.4 m/s - 4.9 m/s^2 * t)\n\n(29.4 m/s)/(4.9 m/s^2) = t\n5.996 sec = t
446442
Vertical components\n\nv = u + at\n0 = u + (-32)(2.1s)\nu = 67.2ft/sec\n\nMaximum height the ball reached\n\nv^2 = u^2 +2gs\n0 = (67.2)^2 + 2(-32)(s)\n64s = 4515.84\ns = 70.56\n\ndownward velocity 0.8 seconds after the ball reached the max.height.\n\nv = u +at\nv = o +(32)(0.8)\nv = 25.6 ft/sec\n\ndistance from the maximum height\n\nv^2 = u^2 +2gs\n(25.6)^2 = 0 + 2(32)s\ns = 10.24 ft\n\nHeight of the fence = 70.56 ft - 10.24 ft\n\n\nHeight of the fence is 60.32 ft...\n\nthere ya go...
405901
h = 4tanθ\nVh = dh/dt = (dh/dθ)(dθ/dt)\ndh/dθ = 4/cos^2θ\ncos^2θ = 4^2/(4^2 + 19^2)\nVh = (16 + 361)/4)(5/361)\nVh = (5/4)(16/361 + 1)\nVh = 1.3054 km/sec
468074
let the height of cliff be h\nthen \ndistance of first ship =x\ndistance of first ship =x+500\nnow \n1.h/x=tan 24\n2.h/(x+500)=tan 15\nfrom 1\nx=hcot24\nfrom 2\nx=hcot15-500\ntherefore\nh(cot15-cot24)=500\nh=336.47
371015
the intercepe on the h axis will give 1/mass of the spring.\nSince, T is directly proportional to m, so if T decreases, this means m decreases, and hence the height of the mass loaded on the spring, from the ground increases.\nWhen T=0, then, the loaded mass must be zero. Hence, the h intercept gives the mass of the spring only.
654374
I'll give you a head start, and we'll see if you can figure it out from there:\n\nFirst, the givens:\n\nd = height from the ground to the point at which the two balls are released = 122.5 meters\n\nv1 = initial velocity of the first ball = 0\nv2 = initial velocity of the second ball\nt1 = time ball 1 is dropping (seconds)\nt2 = time ball 2 is dropping (seconds)\nt2 = (t1)-2 seconds\n\na = gravitational acceleration = 9.8 m/sec^2\n\nFor both balls:\n\nd = vt + at^2\n\nFor the first ball, v1 = 0, so d = at1^2\n\nFor the second ball, d = v2t2 + at2^2\n\nThe distance in both case is the same, so:\n\nat1^2 = d = v2t2 + at2^2\n\nOkay, so here you go:  You have the basic equations,so solving is trivial arithmetic:\n\nd = 122.5 meters = (9.8 m/sec^2) (t1)^2\n\nand solve for t1.  Once you know t1, then you know that t2 is two seconds less, so with that value you can solve:\n\nd = 122.5 meters = v2t2 + (9.8 m/sec^2)(t2)^2\n\nI leave the rest to you!  (And I have calculated the answer if you want to check it)
163378
Assuming no air resistance, the first equation is:\n\n400=1/2 a t^2  (a, as you said, is 32.2 ft/sec^2)\n\nt, the time calculates to be abot 4.98 seconds\n\nthen, v=at, gives v=(32.2)*(4.98)=160 ft/sec\n\nIn the real world, the speed after 400 feet of fall will be less than that due to air resistance, which would slow a brick considerably at 160 ft/sec (over 100 miles/hr)
84350
All objects accelerate downward at a rate of 9.8 m/s/s.  If the object falls 2.6 m, at that rate, it would it the ground in approximately .53 seconds at a speed of 5.2m/s. To calculate it falls approx. 4.9m the first second then continues to speed up at a rate of 9.8m/s/s after that.  Since 4.9m is farther than the distance fallen, divide 2.6 by 4.9 and get the time taken to fall.  Take that answer and multiply that by the constant rate of acceleration to get 5.2m/s