Question

A body is projected at an angle 8 from vertical with
speed u. The velocity at maximum height is
0
u sine
ucose
u​

Answer 1

Answer:

Let the velocity of projectile be u.

Angle of projection with the horizontal θ=60

o

So, initial velocity in horizontal direction u

x

=ucosθ=u×cos60

o

=0.5u

Horizontal distance covered in 5 seconds is 90 m

Thus 0.5u=

5

90

⟹ u=36 m/s

Now the body is projected vertically upwards with a speed u=36 m/s

Maximum height reached H=

2g

u

2

=

2×10

36×36

=64.8 m