# 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)?

2 answers

###### Question:

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)?

## Answers

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)

Answer:h= 78.4 mExplanation: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 Datav₀y= 0t= 4 sg= 9,8 m/s²We apply the formula (1) to calculate vfyvfy= v₀y+g*tvfy= 0+ (9,8)*(4)vfy= 39.2 m/sWe apply the formula (2) to calculate hvfy²=v₀y²+2*g*h(39.2)²=0+2*9.8*hh = (39.2)² / (2*9.8)h= 78.4 m

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