TCE Physics Practice Questions

(a) u_x = 18.0 cos 35.0° = 14.7 m s⁻¹; u_y = 18.0 sin 35.0° = 10.3 m s⁻¹ upward.

(b) Take up as positive, origin at the launch point. The ball lands at s_y = −12.0 m with a = −9.81 m s⁻². Using s = ut + ½at²: −12.0 = 10.3t − 4.905t², so 4.905t² − 10.3t − 12.0 = 0. The quadratic formula gives t = [10.3 + √(10.3² + 4 × 4.905 × 12.0)]/(2 × 4.905) = 2.94 s (the negative root has no physical meaning).

(c) Range = u_xt = 14.7 × 2.94 = 43.3 m. Horizontal velocity at landing is still 14.7 m s⁻¹. Vertical: v_y = u_y + at = 10.3 − 9.81 × 2.94 = −18.5 m s⁻¹ (downward). Speed = √(14.7² + 18.5²) = 23.7 m s⁻¹, at tan⁻¹(18.5/14.7) = 51.4° below the horizontal.

Marking notes: 1 mark per component in (a). In (b): 1 for the correct sign of displacement (−12.0 m), 1 for setting up the quadratic, 1 for 2.94 s. In (c): 1 for range, 1 for speed, 1 for direction stated as below the horizontal. A common error is to find the time to the top (1.05 s) and double it, which ignores the 12.0 m drop. As a check, the maximum height reached is 5.43 m above the platform, so the ball falls 17.4 m from its highest point.