91Ó°ÊÓ

The ball travels before heating the ground when it is thrown horizontally at from a height above the ground.

The following figure shows the motion of the ball thrown horizontally from a height above the ground.

The the height from which the ball is thrown is,

$h=u_{y}t+\frac{1}{2}{a}_y{t}^2$

Here, h is the vertical height from the ground, $u_y$ is the initial speed of the ball, $t$t is the time, $a_y$is the acceleration of the ball due to gravity.

The total horizontal distance travelled by the ball is,

$\begin{array}{l}{s}_h=vt\\ 50.0\text{m}=(25\text{m}/\text{s})t\\ t=2.0\text{s}\end{array}$

The travelling period of ball before heating the ground is 2.0 d.

The height was the ball thrown

role="math" localid="1647475901657" $\begin{array}{l}h=u_{y}t+\frac{1}{2}{a}_y{t}^2\left(\text{l}\right)\\ h=(0\text{m}/\text{s})(2.0\text{s})+\frac{1}{2}(9.8\text{m}/\text{s}){(2.0\text{s})}^2\\ =19.6\text{m}\end{array}$

Therefore, the height was the ball thrown is $19.6\text{m}$.

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