Question

if maximum height = 30√3 and horizontal range= 120m find initial velocity and angle of projection?​

Answer 1

Answer:

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

Answer 2

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