Question

19. A stone dropped from a cliff hits the ground 3.5 seconds later. A second rock is thrown
horizontally from the same cliff and lands 24.5 meters from the base.
a. Find the height of the cliff
b. Find the initial velocity of the thrown rock

Answer 1

For the stone that is dropped from a cliff and hits the ground 3.5 seconds later and for the second thrown rock that lands 24.5 meters from the base of the cliff, we have:

a. The height of the cliff is 60.09 m.  

b. The initial velocity of the thrown rock is 7 m/s.

a. The height of the cliff can be calculated with the following equation:

$y_{f} = y_{i} + v_{i_{y}}t - \frac{1}{2}gt^{2}$  (1)

Where:    

$y_{f}$: is the final height = 0 (when the stone reaches the ground)

$y_{i}$: is the initial height =?

$v_{i_{y}}$: is the initial velocity in the y-direction = 0 (the rock is dropped)

t: is the time = 3.5 s

g: is the acceleration due to gravity = 9.81 m/s²

Hence, the height of the cliff is:

$0 = y_{i} - \frac{1}{2}9.81 m/s^{2}*(3.5 s)^{2}$

$y_{i} = 60.09 m$

Therefore, the height of the cliff is 60.09 m.

b. The initial velocity of the thrown rock can be calculated as follows:

$v_{i_{x}} = \frac{x}{t}$   (2)

Where:

$v_{i_{x}}$: is the initial velocity in the x-direction =?

x: is the distance = 24.5 m

Hence, the initial velocity is:

$v_{i_{x}} = \frac{x}{t} = \frac{24.5 m}{3.5 s} = 7 m/s$

You can find another example here https://brainly.com/question/2140815?referrer=searchResults    

I hope it helps you!