Question

A projectile is thrown horizontally from the top of a tower and strikes the ground after 3s at an angle of 45 with the horizontal. Find the height of the tower and speed with which the body was projected. Given g=9.8 ms-2

Answer 1

Answer:

For the body thrown horizontally

ux=u

uy=0

ay=9.8

t=3

Therefore sy=uyt+12×9.8×32

=44.1m

[Let vv and vx be the vertical & horizotal component when it strikes the ground]

vy=uy+ayt

=0+9.8×3

=29.4m/s

The body strikes the ground at 45∘

tan45=vyvx=>29.4vx=1

Therefore vx=29.4

Therefore the horizontal speed at the time of projection is same as horizontal speed at the time of striking the ground

Therefore u=29.4

Hence d is the correct answer