A car accelerating at the rate of 2 m/ s2 from rest from origin is carrying a man at the rear end who has a gun in his hand. the car is always moving along positive X axis . At t=4 s , the man fires a bullet from the gun and the bullet hits a bird at t = 8 s. The bird haa a position vector 40i + 80j+ 40k . Find velocity of projection of the bullet . Take the Y axis in the horizontal plane. (g =10 m/s2)
Answers
Answer:
Given : Acceleration of car a=2m/s
2
Initial velocity of car u=0
∴ Distance moved by car in 4 s
S
c
=0+
2
1
×2×4
2
=16
i
^
Position vector of bird
r
b
=40
i
^
+80
j
^
+40
k
^
∴ Position vector of bird w.r.t car
r
bc
=
r
b
−
S
c
=24
i
^
+80
j
^
+40
k
^
Let the projection velocity of the bullet in car frame be V making an angle θ with x axis in x-z plane.
⟹ u
x
=Vcosθ and u
z
=Vsinθ
Time taken by bullet to hit the bird t=8−4=4 s
Car frame :
x direction : S
x
=u
x
t where $S_x = 24$$
∴ 24=Vcosθ×4 ⟹Vcosθ=6 .................(1)
z direction : S
z
=u
z
t+
2
1
a
z
t
2
where S
z
=40 and a
z
=−g=−10m/s
2
∴ 40=Vsinθ×4−
2
10×4
2
⟹Vsinθ=30 .................(2)
Squaring and adding (1) & (2) V
2
=30
2
+6
2
=936
⟹ V=30.6m/s (in car frame)
Also from (2) /(1) we get tanθ=5 ⟹θ=tan
−1
5=78.7
o
Explanation:
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