Question

A particle at a height h is projected from the ground with an angle 30 degree from the horizontal it strikes the ground making an angle 45 degree with the horizontal it is again projected from the same point at a height h with the same speed but little angle of 60 degree with horizontal and angle it makes with horizontal when it strikes the ground

Answer 1

Hola

HERE IS THE ANSWER


▶case 1

horizontal initial velocity=horizontal final velocity

u cos30=v cos45............(1)

v=(u cos30)/cos45.....(2)

v^2 sin^245=u^2 sin^230+2gh.......(3)

putting value of v in (3):

2gh=u^2 (cos^2 30-sin^230)....(4)

▶Case 2

horizontal initial velocity=horizontal final velocity

u cos60=v cos  θ............(5)

v=(u cos60)/cos  θ.....(6)

v^2 sin2 θ=u^2 sin^260+2gh.......(7)...as the height remains same.

putting value of v in (8):

2gh=u^2 (tan2 θcos2 60- sin2^60)....(9)

putting 2 gh value from (4) in (9) and hence solving:

tan^2 θ=5

 θ =tan^-1 √5

HOPE IT HELPS YOU