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WAEC 2016 · Paper 2 · Q7

A ball is thrown vertically upwards from ground level and its height after tt seconds is (15.4t−4.9t2) m(15.4t - 4.9t^2)\text{ m}. Find, correct to one decimal place, the:

  1. (a)

    time it takes to reach the maximum height;

  2. (b)

    maximum height it attains.

Worked solution (try it first)

(a)

  1. The ball is momentarily at rest at the top: dhdt=15.4−9.8t=0\dfrac{dh}{dt} = 15.4 - 9.8t = 0.
  2. So t=15.49.8t = \dfrac{15.4}{9.8}
    =117= \dfrac{11}{7}
    ≈1.6\approx 1.6 s.

(b)

  1. Put t=117t = \frac{11}{7} into the height: 15.4×117=24.215.4 \times \frac{11}{7} = 24.2 and 4.9×(117)2=12.14.9 \times \left(\frac{11}{7}\right)^2 = 12.1.
  2. So the maximum height is 24.2−12.1=12.124.2 - 12.1 = 12.1 m.

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