Question

A rock is thrown upward with a velocity of 24 meters per second from the top of a 20 meter high cliff, and it misses the cliff on the way back down. When will the rock be 7 meters from ground level? Round your answer to two decimal places.

Answer 1

Answer: 1.19 s

Step-by-step explanation:

This situation is related to parabolic motion; hence we can use the following equation to find the time the rock is 7 meters from ground level:

$y=y_{o}+V_{o}sin \theta t-\frac{g}{2}t^{2}$ (1)

Where:

$y=20m-7m=13 m$  is the height (position) of the rock when it is 7 meters from ground level (taking into account the initial position is 20 m)

$y_{o}=20 m$  is the initial height of the rock

$V_{o}=24 m/s$ is the rock's initial speed  

$\theta=0\°$ is the angle at which the rock was thrown  (upward or vertically)

$t$ is the time  at whch the rock is in position $y$

$g=9.8m/s^{2}$  is the acceleration due gravity

Isolating $t$ from (1):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (2)

Solving with the given data:

$t=\sqrt{-\frac{2(13 m-20 m)}{9.8m/s^{2}}}$ (3)

$t=1.19 s$ (4) This is the time at which the rock is 7 meters from ground level.