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Question
On lighting a rocket cracker it gets projected in a parabolic path and reaches a maximum height of 4 m when it is 6m away from the point of projection. Finally it reaches the ground 12 m away from the starting point. Find the angle of projection
Solution
Equation of the parabola be x2 = – 4ay .......(1)
B(6, – 4) lies on parabola
62 = – 4a(– 4)
`36/16` = a
⇒ a = `9/4`
(1) ⇒ x2 = `- (9/4) y`
x2 = – 9y .......(2)
Now need tofind slope at (– 6, – 4)
Diff (2) w.r.to x
2x = `- 9 ("d"y)/("d"x)`
`("d"y)/("d"x) = (2x)/(-9)`
At(– 6, – 4), `("d"y)/("d"x) = (2(- 6))/(- 9)`
= `12/9`
= `4/3`
tan θ = `4/3`
θ = `tan^-1 (4/3)`
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