A ball is thrown from the ground to clear a wall $3 m$ high at a distance of $6 m$ and falls $18 m$ away from the wall. Find the angle of projection of ball.

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Solution

In this question we use equation of trajectory to solve this problem
whih can be written as : $y = x tan θ (1 - \frac{x}{R})$
From the figure we can see that it passes through $(x, y) = (6, 3)$ and $R = 6 + 18 = 24$
using these in the equation we get : $3 = 6 tan θ (1 - \frac{6}{24})$
this gives : $tan θ = \frac{2}{3}$
$θ = {tan}^{- 1} \frac{2}{3}$

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A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18 m away from the wall. Find the angle of projection of ball.

In this question we use equation of trajectory to solve this problem whih can be written as : y=xtanθ(1−xR)From the figure we can see that it passes through(x,y)=(6,3) and R=6+18=24using these in the equation we get : 3=6tanθ(1−624)this gives :tanθ=23θ=tan−123

A ball is thrown from the ground to clear a wall $3 m$ high at a distance of $6 m$ and falls $18 m$ away from the wall. Find the angle of projection of ball.