Question

A ball is projected from the ground into the air. At a height of 5 m, its velocity is \( \vec{v} = (5 \hat{i} + 5 \hat{j}) \, \text{m/s} \). The maximum height reached by the ball is: \( (g = 10 \, \text{m/s}^2) \)

• A. 8.75 m
• B. 5.5 m
• C. 6.25 m
• D. 10 m

Hint:

Break 2D projectile into horizontal and vertical components.
Use vertical motion equations: \( v_y^2 = u_y^2 - 2g(h - h_0) \)
Remember: at max height, vertical velocity = 0.

Answer 1

The Correct Option is A

Step 1: Vertical component of velocity at height 5 m is \( v_y = 5 \, \text{m/s} \) Step 2: Use vertical motion equation: \[ v^2 = u^2 - 2g(h - 5) \Rightarrow 0 = 25 - 20(h - 5) \Rightarrow h - 5 = 1.25 \Rightarrow h = 5 + 1.25 = 8.75 \, \text{m} \] Conclusion: \( \boxed{8.75 \, \text{m}} \) is the maximum height.