Question

A rock is thrown upward with a velocity of 21

meters per second from the top of a 22
meter high cliff, and it misses the cliff on the way back down. When will the rock be 9
meters from ground level? Round your answer to two decimal places.

Answer 1

Let's start by using the formula for the height of an object thrown upward:

h(t) = -0.5gt^2 + v0t + h0

where:

h(t) is the height of the object at time t

g is the acceleration due to gravity (9.8 m/s^2)

v0 is the initial velocity (21 m/s)

h0 is the initial height (22 m)

We want to find the time t when the rock is 9 meters from ground level. So, we can set h(t) equal to 9 meters and solve for t:

9 = -0.5(9.8)t^2 + 21t + 22

Rearranging this equation, we get:

0.5(9.8)t^2 - 21t - 13 = 0

Using the quadratic formula, we can solve for t:

t = [21 ± sqrt(21^2 - 4(0.5)(9.8)(-13))] / (2(0.5)(9.8))

t ≈ 2.41 seconds or t ≈ 4.01 seconds

We want the time when the rock is 9 meters from ground level on the way down, so we need to choose the larger value of t. Therefore, the rock will be 9 meters from ground level about 4.01 seconds after it was thrown upward.