Question

A ball is projected upwards with velocity of

200 m/s. Find the speed of the ball at half the

maximum height. Take g = 10 m/s2.
​

Answer 1

Answer:-

Given:-

u = 200m/s.

v = 0m/s.

g = -10m/s^2.

Explanation:-

As the velocities and acceleration is given we should apply third kinematic equation.

${v}^{2} - {u}^{2} = 2as$

${0}^{2} - {200}^{2} = - 2 \times 10 \times h$

$h = \frac{40000}{20}$

$h = 2000 \: metres$

Now,half of the height

H = h/2 = 1000 metres.

Now,applying third kinematic equation.

${v}^{2} - {u}^{2} = 2as \:$

${v}^{2} - {200}^{2} = 2 \times 10 \times 1000$

${v}^{2} = 40000 - 20000$

${v}^{2} = 20000$

$v = \sqrt{20000}$

$v = 100 \sqrt{2}$

or,

$v = 100 \times 1.414 = 141.4$

$\boxed{\boxed{v = 100\sqrt{2}m/s}}$

so,the velocity at half the height is 141.4 m/s or 100√2 m/s.