Chapter 6.1, Problem 4PP

a.

To determine

The ball hang time $t$ .

Answer to Problem 4PP

  $t=2.755\text{ s}$

Explanation of Solution

Given:

Initial velocity of ball, $u=27\text{m/s}$

Angle of projectile, $\theta {\text{=30}}^0$

Formula Used:

First equation of motion:

  $v=u+at$

Where, v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

Calculations:

The projectile is:

  

As $\theta \text{=30}°$ and $u=27\text{m/s}$

So, horizontal and vertical component of velocity will be:

  

  $\begin{array}{l}{u}_x=27\cos {30}^0\\ =27\times \frac{\sqrt{3}}{2}\\ =23.38\text{m/s}\end{array}$

And

  $\begin{array}{l}{u}_y=27\sin {30}^0\\ =27\times \frac{1}{2}\\ =13.5\text{m/s}\end{array}$

Now, at the end of the projectile the vertical component of final velocity will be equal but opposite to the vertical component of initial velocity.

i.e. $v_y=-u_y$

So, using the relation

  $\begin{array}{l}a=\frac{v_y-u_y}{t}\\ -9.8=\frac{-13.5-13.5}{t}\\ t=\frac{-27}{-9.8}\\ =2.755\text{ s}\end{array}$

Conclusion:

The ball hang time $t$ is $2.755\text{ s}$ .

b.

To determine

Ball’s maximum height.

Answer to Problem 4PP

  $d_y=9.3\text{m}$

Explanation of Solution

Given:

Initial velocity of ball, $u=27\text{m/s}$

Angle of projectile, $\theta {\text{=30}}^0$

Formula Used:

  $v^2=u^2+2as$

Where, u is the initial velocity, v is the final velocity, a is the acceleration and s is the displacement.

Calculations:

  

At maximum height that is top, the vertical component of velocity is zero.

  $u_y=0$

Now, ball maximum height will be only dependent on vertical motion.

  $\begin{array}{l}{u}_{yf}=0\text{m/s}\\ a_y=g=-9.8{\text{m/s}}^2\\ u_{yi}=13.5\text{m/s}\end{array}$

So, using the relation

  $\begin{array}{l}{u}_{yf}{}^2=u_{yf}{}^2+2a{d}_y\\ \Rightarrow 0^2={13.5}^2+2\left(-9.8\right)d_y\\ d_y=\frac{182.25}{2\times 9.8}\\ =9.3\text{m}\end{array}$

c.

To determine

Horizontal distance travelled by ball before hitting the ground.

Answer to Problem 4PP

  $d_x=64.41\text{m}$

Explanation of Solution

Given:

Initial velocity of ball $u=27\text{m/s}$

Angle of projectile $\theta {\text{=30}}^0$

Formula Used:

Second equation of motion:

  $s=ut+\frac{1}{2}a{t}^2$

Where, u is the initial velocity, t is the time, a is the acceleration and s is the displacement.

Calculations:

As

  $\begin{array}{l}{u}_x=23.38\text{m/s}\\ t=2.755\text{s}\end{array}$

  

So, horizontal distance can be obtained by using the relation

  $\begin{array}{l}{u}_x=\frac{d_x}{t}\\ d_x=23.38\times 2.755\\ =64.41\text{m}\end{array}$

Conclusion:

Horizontal distance travelled by ball before hitting the ground is $d_x=64.41\text{m}$ .