Question:
A projectile needs to be launched such that its maximum height should not be more than 20 m. If the maximum initial velocity of the projectile is 40 m/s, then the horizontal range is nearly
[Acceleration due to gravity = 10 m/s\(^2\)]
A projectile needs to be launched such that its maximum height should not be more than 20 m. If the maximum initial velocity of the projectile is 40 m/s, then the horizontal range is nearly
[Acceleration due to gravity = 10 m/s\(^2\)]
[Acceleration due to gravity = 10 m/s\(^2\)]
Show Hint
Use the vertical velocity to satisfy the maximum height restriction and then calculate horizontal range using horizontal velocity and time of flight.
Updated On: Jun 19, 2026
- 145.6 m
- 118.6 m
- 138.6 m
- 120.6 m
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Maximum height condition.
Maximum height \(H\) is given by: \[ H = \frac{u_y^2}{2g} \le 20 \] where \(u_y\) is vertical component of initial velocity.
Step 2: Vertical component of velocity.
\[ u_y = \sqrt{2 g H} = \sqrt{2 \cdot 10 \cdot 20} = \sqrt{400} = 20~\text{m/s} \]
Step 3: Horizontal component of velocity.
Maximum velocity \(u = 40~\text{m/s}\), so horizontal component: \[ u_x = \sqrt{u^2 - u_y^2} = \sqrt{40^2 - 20^2} = \sqrt{1600 - 400} = \sqrt{1200} \approx 34.64~\text{m/s} \]
Step 4: Time of flight.
Time of flight \(T = \frac{2 u_y}{g} = \frac{2 \cdot 20}{10} = 4~\text{s}\)
Step 5: Horizontal range.
\[ R = u_x \cdot T = 34.64 \cdot 4 \approx 138.6~\text{m} \]
Step 6: Conclusion.
Hence, the horizontal range is approximately 138.6 m.
Maximum height \(H\) is given by: \[ H = \frac{u_y^2}{2g} \le 20 \] where \(u_y\) is vertical component of initial velocity.
Step 2: Vertical component of velocity.
\[ u_y = \sqrt{2 g H} = \sqrt{2 \cdot 10 \cdot 20} = \sqrt{400} = 20~\text{m/s} \]
Step 3: Horizontal component of velocity.
Maximum velocity \(u = 40~\text{m/s}\), so horizontal component: \[ u_x = \sqrt{u^2 - u_y^2} = \sqrt{40^2 - 20^2} = \sqrt{1600 - 400} = \sqrt{1200} \approx 34.64~\text{m/s} \]
Step 4: Time of flight.
Time of flight \(T = \frac{2 u_y}{g} = \frac{2 \cdot 20}{10} = 4~\text{s}\)
Step 5: Horizontal range.
\[ R = u_x \cdot T = 34.64 \cdot 4 \approx 138.6~\text{m} \]
Step 6: Conclusion.
Hence, the horizontal range is approximately 138.6 m.
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