A rock is thrown upward with a velocity of 22 meters per second from the top of a
45 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.

5.76 seconds

Step-by-step explanation:

We are given;

Position of rock from ground level; h(t) = 9 m

Height of cliff;h_0 = 45 m

Velocity;v_0 = 22 m/s

To solve for the time, we will use the gravity formula which is;

h(t) = (-1/2)gt² + (v_0)t + h_0

Plugging in the relevant values to get;

9 = (-1/2)(9.81)t² + 22t + 45

Subtract 9 from both sides to get;

-4.905t² + 22t + 45 - 9 = 0

-4.905t² + 22t + 36 = 0

Finding the roots of the equation through quadratic formula gives;

t = 5.76 or -1.27

t cannot be negative, so we will use the positive value to get;

t = 5.76 s

Thus, the rock will be 9m from the ground level after 5.76 seconds

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