I kick a soccer off a cliff that is 50m, at 15 degrees above horizon at a velocity of 25m/s
a.) what's the distance from the base of the cliff to the place the soccer lands?
b.) what's the speed before it touches the ground (final speed)?

Assuming you have mass of the ball, you can use any College Physics book to find out these answers. You would have to use potential/kinetic energy theorem to find it. But, I am not aware of any way to find it without having ball's mass.

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Originally Posted by shonny7

Assuming you have mass of the ball, you can use any College Physics book to find out these answers. You would have to use potential/kinetic energy theorem to find it. But, I am not aware of any way to find it without having ball's mass.

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I don't see why the mass would be relevant.

You would use a formula that takes in account of the angle of kicikng the, which I don't feel like finding, but the first guy is wrong because the mass is irrelavent just like if you drop a hammer and a feather in a vacuum they would fall at the same time and hit the ground at the same time.

Start with finding the time it takes for the ball to hit the ground going in the up/down (vertical) direction: Velocity in the vertical direction is 25m/s*sin (15 deg) = 6.47 m/s. Find the time it takes for the ball to hit the ground at 50 m below the cliff = -50m.

x= v*t - 9.8 m/s^2 * t^2 ----> -50m = 6.47*t - 9.8m/s^2 *t^2 (this is a quadratic equation) Solve the quadratic equation for the two roots, which happen to be 2.61 and -1.95. You don't care about negative time, so use 2.61 seconds.

Now do the horizontal direction: 25m/s*cos (15) = 24.148m/s
Multiply the speed times the time -----> 24.148m/s * 2.61 = 63m.