Chapter 6, Problem 50A

To determine

To Find: The velocity of the ball when it left the bat.

Answer to Problem 50A

  $u=30.99\text{m/s}$

Explanation of Solution

Given:

Angle of projectile, $\theta ={45}^o$

Distance travelled, $x=98\text{m}$

Gravitational acceleration, $g=9.8{\text{m/s}}^2$

Formula used:

Range of projectile can be calculated as:

  $R=\frac{u^2\sin 2\theta }{g}$

Where,

  $u$ is the velocity

  $\theta$ is the angle of projectile

  $g$ is the gravitational acceleration.

Calculation:

Velocity of the ball can be calculated as:

  $\begin{array}{l}98=\frac{u^2\sin 2\left(45\right)}{9.8}\\ 960.4=u^2\sin 90\\ u=30.99\text{m/s}\end{array}$

Conclusion:

Velocity of the ball when it leaves the bat is $u=30.99\text{m/s}$ .