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Maurinko [17]
3 years ago
12

I throw a ball off the edge of a 15.0m cliff. If I threw it horizontally at 8.0 m/s, how much time did it take to fall?

Physics
1 answer:
pickupchik [31]3 years ago
3 0
Just ignore the horizontal component 

if you have a vertical displacement of 15m, 0ms^1 initial velocity, end velocity is ignored, we know the acceleration due to gravity as 9.81ms^2 so we can work out the time using SUVAT

S=15
U=0
V=?
A=9.81
T=?

S=UT + 0.5 AT^2

UT=0

therefore,

S=0.5AT^2

rearrange to:

T=SQR (2S/A)

T = 1.75 seconds
 


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We know that total displacement of an object from a position x to a position x', d = final position - initial position.

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If we assume the lad's initial position in front of her house is x = 0 m. The lad then moves towards the positive x-axis, 5 m. He then ends up at x' = 5 m. He then finally goes back 6 m.

Since displacement = final position - initial position, and his displacement is d' = -6 m (since he moves in the negative x - direction or moves back) from his initial position of x' = 5 m.

His final position, x" after moving back 6 m is gotten from

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