Question

138. Two balls A and B are thrown with speeds u and u/2 respectively. Both the balls cover the same horizontal distance before
returning to the plane of projection. If the angle of projection of ball B is 15°with the horizontal, then the angle of projection
of Ais :
1
1
-1
sin 1
(4) sin
sin 1
(2)
(3)
(1) sin
F100
2
8
8
3
8
8
to the hori
should a ball be thrown so that its range R is related to the time of flight Tas R = 552? Take
O​

Answer 1

Answer:

Explanation:

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)