b) On lighting a rocket cracker it gets projected in a parabolic path and reaches a maximum height of 4m
when it is 6m away from the point of projection. Finally it reaches the ground 12m away from the starting
point. Find the angle of projection.
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ANSWER
The pictorial representation of the problem is shown below
Let the vertex be at origin. Since the parabola is open downwards its equation is x
2
=−4ay.
It passes through (6,−4).
⇒36=16a
⇒a=
16
36
⇒a=
4
9
Thus the equation of the path becomes x
2
=−4⋅
4
9
y
⇒x
2
=−9y...(1)
Differentiate (1) with respect to x, we get
2x=−9
dx
dy
⇒
dx
dy
=−
9
2x
Hence slope
dx
dy
=−
9
2x
Slope at (−6,−4) is
dx
dy
=−
9
2(−6)
=
3
4
That is, tanθ=
3
4
⇒θ=tan
−1
(
3
4
)
Hence the angle of projection θ is tan
−1
(
3
4
)
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