A question

用戶85377

2007-10-07 12:08 am

Show the steps very clearly.

1. A volleyball is thrown vertically upwards at a height of 2m above the ground. It takes 3s to reach a height of 10m above the releasing point. Take g = 10ms^-2 and the downward direction as positive.

(a) Find the initially velocity of the volleyball.
(b) Find the velocity of the ball at a height 10m below the maximum point.

回答 (1)

用戶36244

2007-10-07 12:48 am

✔ 最佳答案

a) By using the equation

s= ut + (1/2) at^2,
-10 = 3u + 0.5 (10) 3^2
u= (-10- 45)/ 3
u= -55/3 = - 18.3 m/s

The initially velocity of the volleyball is -18.3 m/s.
b)
When the ball is at maximum point, v =0

v^2 - u^2 = 2as
0 - (-55/3)^2 = 20s
s = - 55^2/ 180 = -16.806

The displacement from releasing point to maximum point is -16.806 m
By using the equation

v^2 - u^2 = 2as
v^2 = 2 (10) (-16.806+ 10) + (-55/3)^2
v = 14.1 m/s

The velocity of the ball at a height of 10m below the maximum point is 14.1 m/s.

2007-10-11 23:39:07 補充：
Please note that the motion of the ball is upwards, it is opposite to the direction which downwards is taken as positive, the displacement therefore should be negative.

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