A ball projected vertically upward with a speed of 50 ms-1. Then the speed at half of the maximum height is :(g= 10 m/s2) - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

Since a ball is thrown vertically upward,
Initial velocity= 50 m/s
Final velocity once the ball reaches the maximum height= 0 m/s

we know, $v = u + at$
$\Rightarrow 0 = 50 - 10 t$ (Acceleration is negative since the ball is going against gravity)
$∴ t = 5 s$

$v^{2} - u^{2} = 2 as$
Where s is the displacement = maximum height(h)
$\Rightarrow 0 - 50^{2} = 2 (- 10) h$
$∴ h = \frac{2500}{20} = 125 m$

The speed at the half of maximum height will be:
$v^{2} = u^{2} + 2 as$
$v^{2} = {(50)}^{2} - 2 (10) (\frac{125}{2})$

$v^{2} = 2500 - 1250 = 1250$

$v = \sqrt{1250} \approx 35 m / s$

A ball projected vertically upward with a speed of 50 ms^-1. Then the speed at half of the maximum height is :
(g= 10 m/s²)