Question

A ball is thrown upward and reaches a height of 64 feet, its initial velocity should be (g = 32 ft/sec)
0 64 ft/sec
O 72 ft/sec
O 82 ft/sec
04096 ft/sec​

Answer 1

Explanation:

It is given that,

Maximum height reached by the ball, h = 64 feet = 19.507 m

At maximum height, the final velocity of the ball, v = 0

We need to find the initial velocity of the ball i.e. u. It can be calculated using third equation of motion as :

u = 19.55 m/s

So, the initial velocity of the ball is 19.55 m/s. Hence, this is the required solution.

Answer 2

 Hmax=  u^2/2g

64= u^2/2 X 32

64^2=u^2

u= 64ft/sec

 

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