50. The equations of motion of a projectile are given by 36t m and 2y = 96t – 9.8t2 m. The angle of projection is
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Answer
Correct option is
A
sin
−1
(4/5)
Given: x−36t,2y=96t−9.8t
2
or y=48t−4.9t
2
Let the initial velocity of projectile be u and angle of projection is θ. Then,
Initial horizontal component of velocity,
u
x
=ucosθ=
dt
dx
t=0
=36 or ucosθ=36---(i)
Initial vertical component of velocity,
u
y
=usinθ=
dt
dy
t=0
=48 or usinθ=48---(ii)
Dividing (ii) by (i), we get
tanθ=
36
48
=4/3
∴sinθ=4/5
or θ=sin
−1
(4/5)
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