Question

A stone is thrown at an angle of 34.0° above the horizontal from the top edge of a cliff with an initial speed of 18.3 m/s. A stopwatch measures the stone's trajectory time from the top of the cliff to the bottom at 2.6 s. What is the height of the cliff?

Answer 1

Step-by-step explanation:

Using the formula for calculating the maximum height in projectile

H = u²sin²theta/2g

u is the initial velocity = 18.3m/s

g is the acceleration due to gravity = 9.8m/s²

theta = 34°

Substitute

H = u²sin²theta/2g

H = 18.3²(sin34)²/2(9.8)

H = 334.89(0.5592)²/19.62

H = 5.337m

Hence the height of the cliff is 5.337m