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. 145.6 m
• B. 118.6 m
• C. 138.6 m
• D. 120.6 m

Hint:

Use the vertical velocity to satisfy the maximum height restriction and then calculate horizontal range using horizontal velocity and time of flight.

Answer 1

The Correct Option is C

Step 1: Maximum height constraint.
H = u_y² / (2g) ≤ 20 m.

Step 2: Finding vertical component.
u_y = √(2gH) = √(2·10·20) = √400 = 20 m/s.

Step 3: Determining horizontal component.
Total u = 40 m/s, so u_x = √(u² - u_y²) = √(1600 - 400) = √1200 ≈ 34.64 m/s.

Step 4: Time of flight.
T = 2u_y/g = 2·20/10 = 4 s.

Step 5: Horizontal range.
R = u_x · T = 34.64 × 4 ≈ 138.6 m.

Step 6: Conclusion.
The horizontal range is approximately 138.6 m.