Question

A ball is projected with velocity U at an angle A with horizontal plane. It's speed when it makes an angle B with the horizontal is

Answer if u know
irrelevant answer will be reported​

Answer 1

You must have given if the rangeo of projectile remains same in both cases

OR

the maximum height remains same

• taking range as same in both cases

we get

$r = u {}^{2} \sin(2a) \div g = x {}^{2} \sin(2b) \div g$

$u {}^{2} \sin(2a) = x {}^{2} \sin(2b)$

$u { }^{2} \div x {}^{2} = \sin(2b) \div \sin(2a)$

$x = u \div \sqrt{ \sin(2b) \div \sin(2a) }$

• if maximum height is same

$h = u { }^{2} sin ^{2} (a) \div 2g = x {}^{2} sin^{2}(b) \div 2g$

usin(a)=xsin(b)

therefore

x=usina÷sinb

IF QUES.WERE COMPLETE ANS WOULD HAVE BEEN MORE NEARER

HOPE IT HELPS