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 wa s the cliff (rounded to the nearest tenth of a meter)?
2 answers:
<span>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)</span>
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+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
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