Question

A stone is thrown horizontally with an initial speed of 8 m/s from the edge of a cliff. a stop watch measures the stone's trajectory time from the top of the cliff to the bottom to be 4.3 s. what is the height of the cliff?

Answer 1

Answer:

90.6m

Step-by-step explanation:

Initial speed = 8m/s

Time = 4.3 s

In studying the motion of an object, we often make use of kinematic equation

s = ut + 1/2at^2

v^2 = u^2 + 2as

a = (v - u) / t

v = final velocity

u = initial velocity

s = distance or height

t = time

a = acceleration due to gravity

Given time to hit the ground as 4.3s, we calculate the height by using

s = ut + 1/2at^2

Since the stone is moving against gravity, a is negative

u = 0

s = 0(4.3) +1/2(-9.8)(4.3)^2

s = -90.6m

The height of the cliff is 90.6m