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# Answer to Question #128774 in Mechanics | Relativity for Marijke

Question #128774
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.
1
2020-08-12T16:36:52-0400

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.

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