Aball is thrown horizontally from a cliff and hits the ground 4 seconds later 40 meters from the base of the cliff. how high was the cliff (rounded to the nearest tenth of a meter)?

h= 78.4 m

Explanation:

The ball moves with a uniformly accelerated movement in the vertical direction (y), we apply the following formulas:

vfy= v₀y+g*t Formula (1)

vfy²=v₀y²+2*g*h Formula (2)

h: hight in meters (m)    

t : time in seconds (s)

v₀y: initial speed in y  (m/s)  

vfy: final speed in y ( m/s   )

g: accelerationdue to gravity  (m/s² )

Known Data

v₀y= 0

t= 4 s

g= 9,8 m/s²

We apply the formula (1) to calculate vfy

vfy= v₀y+g*t

vfy= 0+ (9,8)*(4)

vfy= 39.2 m/s

We apply the formula (2) to calculate h

vfy²=v₀y²+2*g*h

(39.2)²=0+2*9.8*h

h = (39.2)²  /  (2*9.8)

h= 78.4 m

pe = weightxheight = 2000x1000=2,000,000joules