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

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

Solution:

$velocity of the particle at half of the maximum height is v^{2} - u^{2} = 2as v^{2} - u^{2} = 2 (- g) . \frac{u^{2}}{4g} v^{2} - u^{2} = - \frac{u^{2}}{2} v^{2} = u^{2} - \frac{u^{2}}{2} v^{2} = \frac{u^{2}}{2}$

$v = \frac{u}{\sqrt{2}} = \frac{50}{\sqrt{2}} [g = 10 {ms}^{- 2}] = 35.3 v = 35 m/s$

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

Solution:

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