Physics · Uniformly accelerated motion, velocity-time, position-time graph, relations for uniformly accelerated motion, relative velocity
A boy throws balls into air at regular interval of 2 second. The next ball is th
A boy throws balls into air at regular interval of 2 second. The next ball is thrown when the velocity of first ball is zero. How high do the ball rise above his hand? [Take \( \left.\boldsymbol{g}=\mathbf{9 . 8} \boldsymbol{m} / \boldsymbol{s}^{2}\right] \)
- A. \( 4.9 \mathrm{m} \)
- B. \( 9.8 m \)
- C. \( 19.6 m \)
- D. 29.4 \( m \)
Step-by-step solution
The time between throws is 2 s, and the next ball is thrown when the first ball's velocity becomes zero at its highest point. Thus, time to reach maximum height is 2 s. Using v = u - gt, with v=0, we get u = gt = 9.8*2 = 19.6 m/s. Maximum height h = u^2/(2g) = (19.6^2)/(2*9.8) = 19.6 m.
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