# Part A With what minimum speed must you toss a 100-g ball straight up to just touch the 15-m-high...

## Question:

Part A

With what minimum speed must you toss a 100-g ball straight up to just touch the 15-m-high roof of the gymnasium if you release the ball 1.5 m above the ground? Solve this problem using energy.

Part B

With what speed does the ball hit the ground?

## Energy conservation:

This law states "In the isolated conditions the total energy of a system always remains conserved. It may change one form to the other."

Generally, two types of energies are considered:

(a) Kinetic energy:

It is the energy of a moving object.

The kinetic energy is formulated as:

$$\boxed{\text{KE}=\frac{1}{2}mv^2}$$

Here:

• {eq}m {/eq} is the mass of the object.
• {eq}v {/eq} is the velocity of the object.

(b)Potential energy:

Under the gravitational conditions, the energy stored due to height is called the potential energy.

The potential energy is formulated as:

$$\boxed{\text{PE}=mgh}$$

Here:

• {eq}m {/eq} is the mass.
• {eq}g {/eq} is the acceleration due to gravity.
• {eq}h {/eq} is the height.

Given:

• The mass of the ball {eq}\Rightarrow m=100 \ \text{g}=0.1 \ \text{kg} {/eq}
• The height of the gymnasium roof {eq}h=15 \ \text{m} {/eq}
• The height from where the ball is thrown {eq}h_1=1.5 \ \text{m} {/eq}

It is just touching the roof thus the final velocity of the ball is zero.

(a) From the energy conservation:

{eq}\Rightarrow (\text{KE+PE})_{\text{initial}}=(\text{KE+PE})_{\text{final}} {/eq}

{eq}\begin{align} \Rightarrow \frac{1}{2}mu^2+mgh_1=0+mgh \end{align} {/eq}

Thus the minimum speed given to the ball is:

{eq}\begin{align} \Rightarrow u&=\sqrt{2g(h-h_1)}\\ &=\sqrt{2(10)(15-1.5)}\\ &=\color{blue}{16.432 \ \text{m/s}}\\ \end{align} {/eq}

(b) Now the initial condition is the h height.

From energy conservation:

{eq}\Rightarrow (\text{KE+PE})_{\text{initial}}=(\text{KE+PE})_{\text{final}} {/eq}

{eq}\begin{align} \Rightarrow 0+mgh=\frac{1}{2}mv^2+0 \end{align} {/eq}

Thus the velocity of the ball when it is hitting the ground is:

{eq}\begin{align} \Rightarrow v&=\sqrt{2g(h)}\\ &=\sqrt{2(10)(15)}\\ &=\color{blue}{17.320 \ \text{m/s}}\\ \end{align} {/eq} 