JPJC 2026 Projectile Assignment Solutions

Kinematics Assignment 2026/JPJC/H2 Physics 1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS PROJECTILE MOTION SOLUTIONS 1(a) time taken = 50/10 = 5.0 s 2 21 1Height of cliff = (9.81)(5)2 2 123 m gt   [1] for time [1] for correct substitution and correct and answer Vertical velocity = 9.81 x 5 = 49.05 ms-1 2 2 1 Magnitude of velocity = 10 49.05 50.1 ms   49.05tan 10 78.5 below the horizontalo     [1] vertical velocity [1] for correct calculation of v [1] for correct calculation of angle 2(a) 2 2 2 2 1 2 12.44 0.200 20.0(sin 37 ) (9.81) 2 4.905 12.04 2.24 0 12.04 12.04 4(4.905)(2.24) 2(4.905) 12.04 10.05 9.81 2.25 s or 0.203 s Possible distance = 2.25(20.0)cos(37 ) and 0.203(20.0)cos(37 o s ut at t t t t t t t              ) = 35.9 m or 3.24 m = 35.9 m (nearer the centre of field) [1] for correct substitution of s using vertical component [1] for correct t, using larger t [1] for correct substitution [1] for correct distance. (b) 2 2 2 2 0 [20.0sin(37 )] 2(9.81) 7.38 m v u as s s       Height = 7.38 + 0.2 = 7.58 m [1] for correct substitution [1] for correct answer

Kinematics Assignment 2026/JPJC/H2 Physics 2 (c) 2 2 2 2 1 (20.0sin 37 ) 2(9.81)(2.44 0.20) 10.046 (20.0cos37 ) 10.046 18.9 m s y y v v v           10.046tan 20.0cos37 32.2 below the horizontal.       [1] for correct substitution of values [1] for correct magnitude of v [1] for correct angle. (d) [1]correct shape [1] lower height and range, steeper on the fall. (e) -g [1] Constant a= - g [1] Larger than |g| and less than |g| 0 a t 0