A ball is thrown so that its initial vertical and horizontal components of veloc

Question

A ball is thrown so that its initial vertical and horizontal components of velocity are 40 m/s and 20 m/s, respectively. Use a motion diagram to estimate the ball's total time of flight and the distance it traverses before hitting the ground.

How does the magnitude of the velocity vector at impact compare with the magnitude of the initial velocity vector?

They are the same since the ball during its flight has an upward acceleration as its height increases and a downward acceleration on its way down.

It is greater initially.

They are the same since the magnitude of the vertical component of velocity is the same at each height on the way up and on the way down.

It is greater at impact.

Explanation / Answer

1)
Time of flight, T = 2*Vy/g

= 2*40/9.8

= 8.16 s

2)Range = T*Vx

= 8.16*20

= 163.27 m

3) a) They are the same since the ball during its flight has an upward acceleration as its height increases and a downward acceleration on its way down.

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