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Bingel [31]
3 years ago
7

A ball is released from a tower at a height of 100 meters toward the roof of another tower that is 25 meters high. The horizonta

l distance between the two towers is 20 meters. With what horizontal velocity should the ball be imparted so that it lands on the rooftop of the second building?
Physics
1 answer:
I am Lyosha [343]3 years ago
7 0
-- During the time the ball is flying from the high roof to the low roof,
it's going to fall (100-25) = 75 meters.
How long does it take an object dropped from rest to fall 75 meters ?

Distance = (1/2) · (gravity) · (time)²

75 m = (4.9 m/s²) · (time)²

Time² = (75 m) / (4.9 m/s²)
Time² = 15.31 sec²
Time = √(15.31 sec²)  =  3.91 seconds 

So the ball has to cover the horizontal distance of 20 meters 
in 3.91 seconds.

Distance = (speed) · (time)

20 m = (speed) · (3.91 sec)

Speed = (20 m) / (3.91 sec)

Speed  =  5.11 m/s
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Answer:

20.96 m/s

Explanation:

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y = uᵧt + gt²/2

Since the puck slides off horizontally,

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For the horizontal component of the motion

x = uₓt + gt²/2

x = horizontal distance covered by the puck

uₓ = horizontal component of the initial velocity = 20 m/s

g = 0 m/s² as there's no acceleration component in the x-direction

t = 0.639 s

x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m

For the final velocity, we'll calculate the horizontal and vertical components

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g = 0 m/s²

vₓ = uₓ = 20 m/s

Vertical component

vᵧ² = uᵧ² + 2gy

vᵧ² = 0 + 2×9.8×2

vᵧ = 6.26 m/s

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