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# A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (i) the maximum height to which it rises, (ii) the total time it takes to return to the surface of the earth.

Answer :

(i) Initial velocity = 49 m/s

Final velocity = 0 m/s

$$a \ = \ g \ = \ –9.8 m/s^2$$

We know that, $$v^2 – u^2 = 2 g s$$
$$\Rightarrow 0^2 – (49)^2 = 2 (–9.8) × s$$
$$\Rightarrow s = \frac{49 × 49}{2 × 98} \$$
$$\Rightarrow \ s = 122.5 m$$

(ii) We know that, $$v = u + gt$$
$$\Rightarrow 0 = 49 + (–9.8) × t$$

$$\therefore$$ $$t = \frac{49}{9.8} \$$
$$\Rightarrow \ t = 5 s$$

Total time taken to return the surface of the earth by the ball is 5 s + 5 s = 10 s.