Question

A student competing in a rock slinging contest propels a rock with an initial speed of 22 m/s at an angle of 40.0° above the horizontal from the top of a high cliff. If the rock lands 78 m from the bottom of the cliff, how high is the cliff? Assume no air resistance.

Answer 1

As per the question ,a rock is thrown at an angle 40° above the horizontal from a cliff with an initial velocity 22m/s.

Let initially take the horizontal projectile motion upto the height of cliff..

u=22m/s

$\theta$=40°

g=9.8m/$s^2$

Let r be the range of projectile in its first motion

using r=$\frac {(u^2\times Sin2\theta)}{g}$

r=$\frac {22^2\times Sin80}{9.8}$

on solving we get range

r=47.27m

Now the rock reach at the level pf height of cliff Assume It will be in lateral horizontal projectile.

Height of lateral horizontal projectile=Distance of rock

from the bottom of the cliff - the range in horizontal projectile motion

= 78-47.27

=30.73m

for Height

using ,

H=$\frac {u^2\times Sin^2\theta}{2g}$

30.73=$\frac {u^2\times Sin40}{2\times 9.8}$

solving for u

$u^2$= $\frac {30.73\times 19.6}{Sin^240}$ .....equation 1(say)

Let us calculate half the range

of lateral horizontal projectile

$\frac {R}{2}$=$\frac {u^2\times Sin2\theta}{2g}$

=$\frac {30.73\times 19.6\times Sin80}{2\times 9.8\times Sin^2 40}$

=73.24m

This would be the height of cliff.

Height of the clip should be around 73.24m.