Question

If the initial velocity of a projectile be doubled. Keeping the angle of projection same, the
maximum height reached by it will …………………

Answer 1

hello friend...!!

• we know,

$h(max) \: \: = \: \frac{ {u}^{2} {sin \alpha }^{2} }{2g}$
according to the question,

the angle or projection remains same but,
the initial velocity gets doubled

therefore,

$h(max1) \: \: = \frac{ {(2u)}^{2} {sin \alpha }^{2} }{2g}$

implies,

$h(max1) = \frac{4 {u}^{2} {sin \alpha }^{2} }{2g}$

therefore,

H(max1) = 4 H(max)


implies that, the maximum height becomes 4 times the original height.

______________________________

Hope it is useful...!!!

Answer 2

Answer:BEE QUADRUPLED

Explanation:hello friend...!!

• we know,

according to the question,

the angle or projection remains same but,

the initial velocity gets doubled

therefore,

implies,

therefore,

H(max1) = 4 H(max)

implies that, the maximum height becomes 4 times the original height.

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