Question

A ball thrown by one player reaches the other in 2 seconds. The maximum height attained by the ball above the point of projection will be?(g=10m/s²)

Answer 1

A ball thrown by one player reaches the other in 2 seconds.

so, times of flight , T = 2 sec

Let ball is thrown with speed v which makes an angle $\theta$ with horizontal.

so, horizontal component , $v_x=vcos\theta$

and vertical component, $v_y=vsin\theta$

total displacement in vertical direction = 0.

so, 0 = $v_yT-\frac{1}{2}gT^2$

or, $v_y(2)=\frac{1}{2}\times10\times(2)^2$

or, $v_y=10$ m/s ......(1)

so, maximum height attained by ball, H = $\frac{v^2sin^2\theta}{2g}$

= $\frac{(vsin\theta)^2}{2g}$

= $\frac{v_y^2}{2g}$

from equation (1),

= 10²/2(10)

= 5m

hence, maximum height attained by ball is 5m

Answer 2

Explanation:

pls refer above for the answer