A ball is thrown hor. at 8.0 m/s from a cliff 64m high. How far from the base of the cliff does it land?

Question

I know this is not a very hard physics problem but I'm having some trouble figuring out the answer. Does anyone know how to do this?

Answer 1

The usual strategy in problems like these is to break the motion into components (horizontal and vertical), solve one dimension for the time, and then use that time in the other dimension. In this case, we'll begin with the vertical axis.

The equation of motion is

y - yo = voy t - 1/2 g t^2

where y - yo is the vertical distance traversed (64 m), voy is the initial vertical velocity (zero), g is the local acceleration of gravity (9.8 m/s^2), and t the elapsed time. You know everything except time so find t.

In the horizontal dimension, the motion is described by

x - xo = vox t

where x - xo is the horizontal distance traveled (what we're looking for), vox is the initial horizontal velocity (8 m/s), and t is the elapsed time (what we just found). Solve for x - xo.

Answer 2

We need to find how long it will be before the ball hits the ground. 64=1/2(9.8)t^2 or
t^2=13 so t=3.6 seconds.

Distance from the cliff = 8(3.6) = 28.9 meters