a projectile is fired vertically upwards from ground. Sometime later the projectile is 1000m above the ground and is moving downward with velocity 50m/s. How long has the projectile been in air?(take g= 10m/s^2)
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Answer:
Range of projectile R=
2g
u
2
sin2θ
---- (1)
For maximum height, angle of projection should be 45
o
Maximum height, H=
2g
u
2
sin
2
θ
---- (2)
(2)÷ (1) =
(
g
u
2
sin2θ
)
(
2g
u
2
sin
2
θ
)
=
2Sin2θ
sin
2
θ
=
R
H
=
4
1
H=R×
4
1
Since R= 1000m
H = 250m
Explanation:
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