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3 years ago

A bowling ball is dropped from a height of 24 feet.

Physics

1 answer:

sdas [7]3 years ago

7 0

Answer:

a. H=(1/2)*a*t^2

B.h=16.09ft

C.t= 1.22s

Explanation:

Explanation: Newton's equation of motion. This is called motion under free fall.

S=ut +(1/2)*a*t^2

Let S=H, and a=g

H=ut +(1/2)*g*t^2

H=distance covered in ft(24ft)

u= initial velocity (ft/s)

t= time covered in seconds

g=acceleration due to gravity(ft/s^2)

a=g (32.185ft/s)

When u=0

H=(1/2)*g*t^2

2. H=(1/2)*g*t^2

g=32.185ft/s^2

t=1s

Juxtaposing the values into the equation

H= (1/2)* 32.185*(1)^2

H=16.09ft

3. When the maximum distance covered is 24ft, then using

H=(1/2)*a*t^2

24=0.5*32.165*t^2

24=16.09t^2

1.491=t^2

Find the square root of both sides

t=1.22s

It takes 1.22s to hit the ground

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