poora solve krke bhejna
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Verified 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