Question

A cannon on a cliff 55 m above the ground fires a shell with a velocity of \(50\hat{i}+50\hat{j}\) ms\textsuperscript{−1}. The displacement vector of the shell when it hits the ground is (Acceleration due to gravity = 10 ms\textsuperscript{−2})

• A. \((550\hat{i}-55\hat{j})\)\,m
• B. \((550\hat{i}-500\hat{j})\)\,m
• C. \((500\hat{i}-55\hat{j})\)\,m
• D. \((500\hat{i}-550\hat{j})\)\,m

Hint:

Solve vertical motion for time of flight, then use horizontal velocity (no horizontal acceleration) to get range/displacement.

Answer 1

The Correct Option is A

• Vertical motion: initial y = 55 m, \(u_y = 50\) m/s, \(g=10\) m/s\textsuperscript{2} downward. Solve \(55 + 50t - 5t^2 = 0\).
• This gives \(t = \dfrac{50 + \sqrt{50^2 + 4\times5\times55}}{10} = \dfrac{50 + 60}{10} = 11\) s (positive root).
• Horizontal displacement: \(x = u_x t = 50 \times 11 = 550\) m.
• Vertical displacement (change) = final y - initial y = \(0 - 55 = -55\) m → represented as \(-55\hat{j}\).
• Hence displacement vector = \((550\hat{i} - 55\hat{j})\) m → option (1).