Question

an object is projected horizontally from the top of a tower of height H the line joining the point of projection and point of striking of ground make an angle 45 degree with the ground then with what velocity the objects strike the ground​

Answer 1

▶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

Answer 2

Answer:

hope the answer is correct

Explanation: