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Question: A ball is projected with a velocity 20 ms<sup>–1</sup> at an angle to the horizontal. In order to ha...

A ball is projected with a velocity 20 ms–1 at an angle to the horizontal. In order to have the maximum range, its velocity at the highest position must be –

A

10 ms–1

B

14 ms–1

C

28 ms–1

D

Zero

Answer

14 ms–1

Explanation

Solution

For maximum range q = 450

Hence velocity at highest point = u cos 450

= 20 ×12\frac{1}{\sqrt{2}}= 14 m/s.