Question

A body projected at certain angle (\(\neq 90^\circ\)) from the ground crosses a point in its path at a time of 2.3 s and from there it reaches the ground after a time of 5.7 s. The maximum height reached by the body is (Acceleration due to gravity = 10 ms\(^{-2}\))

• A. 80 m
• B. 120 m
• C. 40 m
• D. 160 m

Hint:

For any two points on a projectile's trajectory at the same height, the time taken from launch to the first point and the time taken from the second point to landing are equal. The problem implies the body is at the same height at \(t=2.3\)s on the way up and some time \(t'\) on the way down. The time from \(t'\) to landing (5.7s) is equal to the time from launch to \(t=2.3\)s. However, the interpretation used in the solution is more direct.

Answer 1

The Correct Option is A

Let the point be crossed at time \(t_1 = 2.3\) s.
The time taken to reach the ground from that point is \(t_2 = 5.7\) s.
The total time of flight, T, is the sum of the time to reach the point and the time from the point to the ground.
\( T = t_1 + t_2 = 2.3 + 5.7 = 8.0 \) s.
For projectile motion, the time of flight is given by \( T = \frac{2u_y}{g} \), where \(u_y\) is the initial vertical component of velocity.
From this, we can find the initial vertical velocity: \( u_y = \frac{gT}{2} \).
\( u_y = \frac{10 \times 8.0}{2} = 40 \) m/s.
The maximum height reached by a projectile, H, is given by the formula \( H = \frac{u_y^2}{2g} \).
Now, substitute the values we found.
\( H = \frac{(40)^2}{2 \times 10} = \frac{1600}{20} = 80 \) m.