Question

16. An object is thrown vertically upwards from the top of a building 20 m high. If the object passes the point it was thrown from after 4 seconds, find: (a) the velocity at which the object was thrown; (b) the time taken when the object is 10 m above the level it was thrown; (c) the time taken when the object is 10 m below the level it was thrown; (d) the velocity with which the object hits the ground. [Take g = 10 ms⁻²]

Accepted Answer

(a) Since it returns to the throwing point after 4s, time to reach maximum height = 2s. At max height v=0: 0=u-g(2) → u=20 ms⁻¹.
(b) Using s=ut-½gt² with s=10: 10=20t-5t² → 5t²-20t+10=0 → t²-4t+2=0. Solving: t = 2±√2, giving t≈3.414s or t≈0.586s.
(c) Using s=-10 (below throwing point): -10=20t-5t² → 5t²-20t-10=0 → t²-4t-2=0. Solving: t = 2±√6, giving t≈4.45s (taking the positive root).
(d) The ground is 20m below the throw point, so s=-20: v²=u²+2gs = 20²+2(10)(20) [taking downward as positive with magnitudes] = 400+400=800, so v=√800=20√2 ≈ 28.28 ms⁻¹ (downwards).

WAEC Further Mathematics 2025 Theory — Question 16

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