A ball is projected from a cliff 96 m high and lands 7.0 s later, a net 96 m below the launch point (air resistance negligible, $g = 9.81$ m s $^{-2}$ ). Using $s = ut - \frac{1}{2}gt^2$ with $s = -96$ m and $t = 7.0$ s (taking upward as positive), calculate the initial vertical velocity $u$ of the ball.

Model answer

20.6 m/s

Also accepted: 21 m/s, 20,6 m/s

Explanation: Substituting into

s=ut-\frac{1}{2}gt^2

-96 = 7u - \frac{1}{2}(9.81)(49)

, so

7u = 240.345-96=144.345

, giving

u \approx 20.6

m s

^{-1}

upward (rounds to about 21 m s

^{-1}

Derived from ZIMSEC Physics Paper 3, November 2003, Q1

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