A ball 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)?

A ball is thrown horizontally from a cliff at a speed of 15 m/s and strikes the ground 45 meters from the base of the cliff. How high was the cliff (rounded to the nearest meter)

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valeriagonzalez0213 valeriagonzalez0213 · Answer 2 · 2019-10-14T16:33:02-04:00

Answer:

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+gt Formula (1)

vfy²=v₀y²+2gh 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+gt

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

vfy= 39.2 m/s

We apply the formula (2) to calculate h

vfy²=v₀y²+2gh

(39.2)²=0+29.8h

h = (39.2)² / (29.8)

h= 78.4 m