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(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.