How do you solve this problem? Bob shoots an arrow straight up with an initial velocity of 288 ft/sec from a h
924813
(288^2)/(2*32.2) is the maximum height.\nat time 288/32.2 seconds\nit hits ground at time time   t+576/32.2, where t is obtained from\n45=288+.5*32.2*t^2
987748
1.Compute initial velocity V0\nUsing equation S=V0*t-(g*t^2)/2 we find that the initial velocity V0=35/3 + 3g/2\n2.What ever goes up must come down V0*t=-(g*t^2)/2 t=2*V0/g\n3.Total time travel = 2t \nYou to the math\n\nBest\n\nHint:  the arrow will stop midair in about 5.3 seconds
162313
Let V = speed (the magnitude of the initial velocity)\na = angle, t = time, \nVx = horizontal (x) component of velocity\nVy = vertical (y) component of velocity\nAx = acceleration in x direction\nAy = acceleration in y direction\nDx = net distance traveled in x direction\nDy = net distance traveled in y direction\n\nVx = Vcos(a)\nVy = Vsin(a)\n\n  Dx = 1/2 Ax*t^2 + Vx*t\nbut Ax = 0, so Dx = Vx*t, t = Dx/Vx\n\n  Dy = 1/2 Ay*t^2 + Vy*t\n(Ay = -g, the acceleration of gravity, negative for downward)\n -1/2*g*Dx^2/(Vcos(a))^2 + Vsin(a)*Dx/(Vcos(a)) = Dy\nMultiply by V^2, and collect terms with V^2 on the left:\n(Dx*tan(a)-Dy) V^2 = 1/2*g*(Dx)^2/(cos(a))^2\nV = sqrt(1/2*g*(Dx)^2/(cos(a))^2/(Dx*tan(a)-Dy))\nsubstitute a = 30 degrees, Dx = 6.05 m, Dy = 3.05-2.2 = 0.85 m, g = 9.8 m/s^2\nV = 9.51 m/s\nFor a = 60 degrees,\nV = 8.63 m/s
349891
Maybe not mathamatics directly but it has a lot to do with Physics. The gravitational pull of the ground, the velocity with which the arrow is let go. Also a little bit of elevated angel to try and hit the target.
923403
Start off by drawing the picture. You have a right triangle, with a hypotenuse of 25, a base (b), and height (h). \n\nDon't get jumbled down by numbers yet, ignore them. Visualize what you are calculating.\n\nVelocity / speed is the derivitive of position. (v = delta x / delta t). So all you need to do is find an eqation relating the geometry using known and unknown quantities and then differentiate with respect to time. Sounds confusing but all you have to do is....\n\nYou know, a^2+b^2 = c^2 for a right triangle. In our problem,\n\nb^2+h^2 = 25^2\n\nThis is our position equation. Now differentiate with respect to time.  DO NOT forget the CHAIN RULE! <------ VERY important\n\n2*b*db/dt +2*h*dh/dt = 0.\n\nFrom here just use algebra; where,\n\nb=7 ft, and db/dt = 2 ft/sc.\n\nWhich leaves two equations and two unknowns.\n\nThe important thing to remember is to take this step by step and try to visualize what you are calculating. Also, looking at the units of the given quatitities can give you some idea of what you will be required to do. \n\nAlso, just remeber that speed / velocity is the derivative of position with respect to time (the units reflect this).
799812
The maximum height is given by the point where the slope (derivative) is zero.\n\nh'(t) = -32t + 96 = 0\n32t= 96\nt = 3\n\nSo the maximum point will be at time t = 3.\n\nh(3) = 5 + 96(3) - 16(3)²\nh(3) = 5 + 288 - 16*9\nh(3) = 293 - 144\nh(3) = 149 ft.\n\nNow just don't stand directly underneath this when the arrow returns to Earth!
766563
though instant the bird leaves the swing we assume that the centre of gravity does not change of the system. Therefore 52*2=153*x, where x represents the swing velocity= 0.6797m/s.\n\nnow the energy for the swing is going to be conserved.  Just a great kinetic energy with potential energy and get the height. mgh=1/2mv2\n\nCalculate h.  That is the answer.\n\nIt took me quite a long time to type all this....so I hope you will give me the best answer.
703199
It is an exercise in elementary calculus to show that (ignoring air resistance) the elevation angle for maximum distance is 45 degrees.  The vertical velocity of the egg will be zero at the peak height, and will increase as v = gt^2/2 until it hits the ground.  Energy is conserved, so the vertical velocity when it strikes the ground will match the vertical velocity with which it was launched, and the horizontal velocity will be constant.   This is enough data to solve the whole thing, but there is a shortcut: in the parabolic path at issue, it is always true that the peak height is 1/4 of the horizontal distance.  Hence, the peak height here is 24.225 meters, and that makes it trivial to compute the time and hence the velocity.
916863
I get that the maximum distance that the ship can fire is 6371m, meaning that any distance greater than 3571m from the western shore should be safe.  It may be tempting to assume that the target ship would be safe within a certain minimum distance, where it would be protected by the mountain peak, but the enemy ship could hit any target within the 3571m limit by angling its guns higher than 45 degrees.\n\nEDIT:\nTo start solving our problem, let's figure out the maximum distance that the enemy ship can fire.  It's not hard, and we know we'll have to figure it out anyway.\n\nIf the ship fires its projectile at exactly 45 degrees, the projectile will make a parabolic arc through the air.  At angles of more than 45 degrees, the projectile will fly higher, but it won't go as far (like a pop-up fly in baseball).  At angles less than 45 degrees, the projectile will fly lower, and hit the water with more horizontal motion (like a line-drive in baseball).  At exactly 45 degrees, it will achieve its maximum horizontal distance.  \n\nIn order to figure out how far the projectile flies, we need to break up its motion into horizontal and vertical components.  Imagine a right triangle with a hypotenuse of 250 units and one angle of 45 degrees.  (The other angle will also be 45 degrees, but that doesn't matter for now.)  The line adjacent to our angle will be as long as the projectile's horizontal momentum.  The line opposite the angle will be as long as the projectile's vertical momentum.  We just need a little bit of trigonometry to figure out these numbers.\n\nsin(a) = opposite / hypotenuse;\nsin(45) = Vv / 250;   (Vv represents the vertical velocity)\nVv = sin(45) * 250 = 177m/s;\n\nSince we're using a right triangle with two 45 degree angles, both sides will be the same length.  If we were using a different angle, we could easily solve for the horizontal momentum using the Pythagorean Theorum.\n\nSo, we know that our projectile had a horizontal velocity of 177m/s and a vertical velocity of 177m/s.  Let's figure out how long it would stay in the air.  We know that it's constantly accelerating towards the earth due to the force of gravity.  Our answer lies in the definition of acceleration: acceleration is a change in velocity over time.  Mathematically, that's (a=ΔV / t).  The "Δ" symbol is the Greek letter delta, often used to represent a change in a value.  \n\nTo figure out how long it stayed up, we only need to worry about the projectile's vertical velocity.  We know that it's initial vertical velocity is 177m/s.  We also know that it's vertical velocity at the top of the arc is 0m/s.  The change between these two points in time is (177m/s - 0m/s =) 177m/s.  The only other thing we need to figure out time is the acceleration due to gravity.  On Earth, it's always 9.81m/s^2. So...\n\na = ΔV / t;\n9.81 = 177 / t;\nt = 177 / 9.81 = 18.04 seconds.  That's how long it takes the projectile to reach the top of it's arc.  It will take equally as long to reach the bottom of it's arc, so the total flight time is 36.09 seconds.  \n\nNow we have a simple distance equation.  Since distance is equal to velocity multiplied by time, the projectile's horizontal distance will be equal to its horizontal velocity multiplied by the amount of time that it stayed in the air.\n\nd = 177 * 36 = 6372m.  (This number is a little higher than the first one, because I rounded a little differently.  It doesn't make a huge difference.)  If we take out the distance to the western shore (2500m + 300M), we get a maximum distance of 3572m from the western shore.  Any ship beyond that distance will be safe.  \n\nAll we need to do now is figure out whether or not the enemy ship can fire at that maximum distance.  If the mountain is too high, the ship will have to increase the angle of its guns, thereby losing range.  We can figure out how high our projectile is going to fly with
759333
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m
979449
The rocket travels in a parabolic trajectory (we assume it has a zero burn duration). It is at the peak with 0 vertical velocity when it lands on the cliff edge at s=200 m height. Under 1g deceleration, it takes SQRT(2*s/g) = 6.387 sec to reach this height. The horizontal distance  traveled is 30 m in 6.387 sec for a velocity of 4.607 m/sec.\nI hope you realize that when you impaled the package with the rocket, all the blood plasma, propane and brandy were lost.\n\nEDIT: It would help if you said why it doesn't seem right. But if it isn't right, we must not be trying to get the payload up the cliff, just across the river. Which makes the cliff height another irrelevant parameter, along with the rocket and payload masses, and also means our rocket doen't travel the hypotenuse as you thought. We want a minimum-energy way to fling the stuff across the river (distance S = 30m), which means a 45-degree launch elevation (and still a parabolic trajectory). At launch, vertical velocity Vy = horizontal velocity Vx. To solve for velocity we use two definitions of flight time T: T = 2Vy/g (the time to ascend to peak height and then descend to the ground; of course T also = 2Vx/g), and T = S/Vx (hor. distance/hor. velocity). Then 2Vx^2 = Sg = 30m*9.80665m/s^2, so Vx=Vy=12.128 m/s, and the velocity magnitude = 17.152 m/s. This is the launch and impact velocity. \nOf course if the rocket has more energy than this minimum value, one must change the launch elevation to compensate and impact velocity will be greater. If it has less energy than the minimum, the explorer croaks.\nNow if this version doesn't seem right, maybe we're on top of the cliff and have to propel it downwards and across the river? I'll let you solve that one.\nP.S. Your problem statement is confusing. You say "... then fire the rocket at ground level toward the supplies." Do you mean toward the explorer?
757769
Here's how to solve this.\nThe velocity is a vector, and right-angle components of velocity operate independently.\nThe bullet is fired with 2000 ft/s horizontal velocity but zero vertical velocity. So you know the horizontal and vertical components of velocity at the instant the bullet leaves the barrel.\n\nHowever, once it leaves the gun's barrel, the bullet experiences no horizontal forces, but vertically it is acted upon by gravity.\n\nSo: 1) how long does it take for the bullet to fall 5 ft, starting with zero vertical velocity? (hint: s = 1/2 a t^2\n\n2) Once you figure out how long until the bullet hits the ground, then figure out how far the bullet has travelled horizontally during that time, given 2000 m/s initial horizontal velocity and no horizontal forces.\n\nDoes this help?
415527
First of all, the sum is incompleted.  Is it for maximum height, or for range?  Secondly, let us assume you are asking for maximum height.  Then differentiating the given equation of motion, d = -5(t+1)^2 + 258 with respect to 't', we get dt/dt = -10(t+1).  At maximum height, the velocity is zero. \nTherefore, -10(t+1)=0, or t = -1.  This is absurd time can not become negative once the motion is initiated.  Check up there is something wrong in the problem.
765211
h = -16x² + 48x + 8\n\nNow clearly it is projected from an elevated position and that position is 8 ft above the horizontal ( when x = 0 h = 8)\n\nSo the NEXT time it reaches 8 ft will be the second time it is there (on the way down)\n\n8 = -16x² + 48x + 8\nie 0 = -16x² + 48x\n= 16x(3 - x)\nSo x = 0 or 3\n\nThus the next time it reaches 8 ft is in 3 second
735664
we know m=0.0043kg, Vfinal=304m/s, d=0.86m and Vinitial=0\n\nwhat we need is the value of acceleration ( remember F=ma )\n\nso the formula we need is\n\nvfina^2=viinitial^2 + 2ad\n304^2=0 + 2*0.86*a\na=(304^2)/(2*0.86)\na=53730.23256\n\nso F=m*a=53730.23256*.0043=231.04 N\n\nForce is 231.04 N\n\n*****************************************************\n\nHere's an alterative if you want to double check  \n\nthe kinetic energy (KE)=1/2 *m*(velocity)^2\n\nthe KE of the bullet must equal the Work needed to fire it\n\nKE=0.5*0.0043*304^2 =198.6944 j\n\nWork=F*d=198.6944\n\nso F=(198.6944)/0.86=231.04 N\n\nAgain we get  force = 231.04 N\n\nSo we're pretty sure this is the correct answer
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.
840248
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...