A body is projected with an angle e. The maximum height reached is h. If the time of flight is 4
sec and g =
=10m/s?, then the value of his
Answers
Answered by
10
hmax =u^2/2g
v=u+at
0=u-10*4
u=40
hmax=40^2/2*10
=80
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Answered by
1
Let the body is projected with velocity u and the angle of projection be θ .
Given : Time of flight T=4 s
∴ Time of flight T=
g
2usinθ
⟹ u
2
sin
2
θ=
4
T
2
g
2
.......(1)
Maximum height reached h=
2g
u
2
sin
2
θ
=
2g
4
T
2
g
2
=
8
T
2
g
∴ h=
8
4
2
×10
=20 m
Given : Time of flight T=4 s
∴ Time of flight T=
g
2usinθ
⟹ u
2
sin
2
θ=
4
T
2
g
2
.......(1)
Maximum height reached h=
2g
u
2
sin
2
θ
=
2g
4
T
2
g
2
=
8
T
2
g
∴ h=
8
4
2
×10
=20 m
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