Question:
A ball projected from ground at an angle of \(45^\circ\) just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is :
A ball projected from ground at an angle of \(45^\circ\) just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is :
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Range \(R=4+6=10\) m. Use the trajectory \(y=x\tan\theta\left(1-\frac xR\right)\) at \(x=4\) m.
Updated On: Oct 31, 2026
- 4.4 m
- 2.4 m
- 3.6 m
- 1.6 m
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The Correct Option is B
Solution and Explanation
The ball lands 4 + 6 = 10 m from the point of projection, so the range \(R=10\) m.
Trajectory equation for projection at \(\theta=45^\circ\): \(y=x\tan\theta\left(1-\frac{x}{R}\right)=x\left(1-\frac{x}{10}\right)\).
At the wall \(x=4\) m: \(y=4\left(1-0.4\right)=2.4\) m.
Answer: (2) 2.4 m
Trajectory equation for projection at \(\theta=45^\circ\): \(y=x\tan\theta\left(1-\frac{x}{R}\right)=x\left(1-\frac{x}{10}\right)\).
At the wall \(x=4\) m: \(y=4\left(1-0.4\right)=2.4\) m.
Answer: (2) 2.4 m
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