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soldier1979 [14.2K]
3 years ago
14

A ball is launched from ground level and hits the ground again after an elapsed time of 4 seconds and after traveling a horizont

al distance of 56 m. The only force acting on the ball is gravity. Which of the following statements best describes the magnitude of the horizontal component of the velocity of the ball? a. It is constant the whole time the ball is in free-fall. b. It is smallest when the height of the ball is maximum. c. It is smallest just before the ball hits the ground.
Physics
1 answer:
Mumz [18]3 years ago
6 0

Answer:

a. It is constant the whole time the ball is in free-fall

Explanation:

If we divide the movement on its vertical and horizontal components, and we concentrate on the vertical component, let's call x-component, and analyze Newton's second's law:

\sum\overrightarrow{F_{x}}=m\overrightarrow{a_{x}}

with \sum\overrightarrow{F_{x}}, a_{x} the acceleration on horizontal direction and m the mass of the ball, because the only force acting on the object is gravity that is always vertical, there're not forces on the horizontal direction that means \sum\overrightarrow{F_{x}}=0 and by (1) that implies a_{x}=0 there's not acceleration on horizontal direction.

Because acceleration is the rate at what velocity changes and there's no acceleration, there's no change in velocity, in other words velocity is constant on horizontal direction.

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b) v = 21.34 m/s

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θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

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KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

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