Question

Two balls A and Bare thrown with speed u and u/2 respectively both ball coverd the same horizontal distance if the angle of projectile of the ball B is 15°with respect to horizontal then angle of projection of A is

Answer 1

The time consumed for  an object to be projected and landing is called the time of flight.

This depends on the initial velocity  and the angle of projection of the projectile.

When the projectile attains  a vertical velocity of zero, this is the maximum height attended by it.

The horizontal displacement of the projectile is termed as range of the projectile,

Solution: R=u2sin2θ/g:

As per problem, :(u/2 )2sin 300/g =u2sin2θ/g:

Therefore, u2sin300/4g=u2sin2θ /g: or, u2/8g= u2sin2θ /g: or, sin2θ =1/8 or, 2θ=sin-1(1/8)    or, θ=1/2 sin-1(1/8): Angle of projection of  A:  is 1/2 sin-1(1/8)  

Answer 2

Answer:

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