Question

A body is projected with a speed 25m/sat an angle Ɵ to the horizontal to have a maximum range. What is the velocity in m/sat the highest point?

Answer 1

Explanation:

Here u=30ms

−1

, Angle of projection, θ=90−30=60

∘

Maximum height,

H=

2g

u

2

sin

2

θ

=

2×10

30

2

sin

2

60

∘

=

20

900

×

4

3

=

4

135

m

Time of flight, T=

g

2usinθ

=

10

2×30×sin60

∘

=3

3

s

Horizontal range = R=

g

u

2

sin2θ

=

10

30×30×2sin60

∘

cos60

∘

=45

3

m