Answer:
Velocity with which rock will hit the ground is 167.045 feet.
Step-by-step explanation:
A rock is thrown upward with a speed of 48 feet per second from the edge of a cliff.
So we will calculate the height attained by the rock will be
v² = u² - 2gh [Third equation of vertical motion]
Where v is the final velocity at height and u s the initial velocity f the rock.
For maximum height attained by rock v = 0
0 = 4² - 2 gh
4² = 2 gh
(48)² = 2 × 32 × h
h = ![\frac{48\times 48}{64}=36 feet](https://tex.z-dn.net/?f=%5Cfrac%7B48%5Ctimes%2048%7D%7B64%7D%3D36%20feet)
Now rock is at h = height of cliff from the ground + height attained by rocked when thrown upwards.
height h' = 400 + 36 = 436 feet.
By the formula v² = 4² + 2 gh [ third equation of vertical motion ]
here 4 = 0 and h' = 436 feet.
So v² = ![0+2\times (3.2)(436)](https://tex.z-dn.net/?f=0%2B2%5Ctimes%20%283.2%29%28436%29)
= 27904
v =
feet per second.
Therefore, velocity with which rock will hit the ground is 167.045 feet.