Question

A baseball is thrown horizontally from a height of 9.60 m above the ground with a speed of 32.7 m/s. Where is the ball after 1.40 s have elapsed?The ball is above the ground at a horizontal distance of ____ m from the launch point.

Answer 1

Given,

The height from whcih the baseball was thrown, h=9.60 m

The initial speed of the ball, u=32.7 m/s

The time instant, t=1.40 s

The velocity of an object is the time rate of change of displacement of the object.

After the ball is thrown there will be no horizontal forces acting on the ball and hence there will be no acceleration in the horizontal direction. That is, the velocity of the ball in the horizontal direction will be constant throughout its flight.

Therefore the horizontal distance of the ball from the point of the launch is given by,

`x=ut`

On substituting the known values,

`\begin{gathered} x=32.7*1.40 \\ =45.78\text{ m} \end{gathered}`

Therefore the ball will be above the ground at a distance of 45.78 m from the launch point after 1.40 s of the launch.