A ball is projected vertically upward with a speed of 50ms−1. Then the speed at half of the maximum height is (g = 10 ms−2)

A ball is projected vertically upward with a speed of $50 {ms}^{- 1}$ . Then the speed at half of the maximum height is (g = $10 {ms}^{- 2}$ )

A
$100 {ms}^{- 1}$
B
$125 {ms}^{- 1}$
C
$35 {ms}^{- 1}$
D
$45 {ms}^{- 1}$

Solution:

Initial velocity $u = 50 m/s$ , Acceleration due to gravity $g = {10ms}^{- 2}$

Maximum height $H_{\max} = \frac{u^{2}}{2g} \Rightarrow H α u^{2}$

$∴ \frac{H_{1}}{H_{2}} = \frac{u_{1}^{2}}{u_{2}^{2}}$

$\Rightarrow \frac{H_{1}}{\frac{H_{1}}{2}} = \frac{50 \times 50}{u_{2}^{2}} \Rightarrow 2 u_{2}^{2} = 2500$

$\Rightarrow u_{2}^{2} = 1250 \Rightarrow u_{2} = 25 \sqrt{2}$

$u_{2} = 35.35 {ms}^{- 1}$

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