Question

A stone leaves a slingshot horizontally with an initial speed of 15.2 m/stromaheight 1.1 m above the ground. Determine the stone's:a) total air timeb) maximum heightc ranged) final velocity

Answer 1

Answer:

a) 0.47 s

b) 1.1 m

c) 7.2 m

d) 15.9 m/s

Explanation:

Part a)

To find the total air time, we will use the equation for the vertical movement of the stone, so

$h=v_{iy}t+\frac{1}{2}gt^2$

Where h is the height, viy is the initial vertical velocity and it is 0 m/s, g is the gravity and t is the time.

Replacing h = 1.1 m, viy = 0 m/s and g = 9.8 m/s² and solving for t, we get

$\begin{gathered} 1.1=0t+\frac{1}{2}(9.8)t^2 \\ \\ 1.1=4.9t^2 \\ \\ \frac{1.1}{4.9}=t^2 \\ \\ \sqrt{\frac{1.1}{4.9}}=t \\ \\ 0.47\text{ s}=t \end{gathered}$

Therefore, the total air time was 0.47 s

Part b)

The stone leaves horizontally, so the maximum height is the initial height of 1.1 m

Part c)

The range can be calculated using the following equation

$\begin{gathered} x=vt \\ x=(15.2\text{ m/s\rparen\lparen0.47 s\rparen} \\ x=7.2\text{ m} \end{gathered}$

Therefore, the horizontal distance traveled was 7.2 m

Part d)

First, we will calculate the final vertical velocity as

$\begin{gathered} v_{fy}=v_{iy}+gt \\ v_{fy}=0+9.8(0.47) \\ v_{fy}=4.6\text{ m/s} \end{gathered}$

Then, the final vertical velocity was 4.6 m/s and the horizontal velocity was 15.2 m/s, so the final velocity was

$\begin{gathered} v_f=\sqrt{15.2^2+4.6^2} \\ v_f=15.9\text{ m/s} \end{gathered}$

Therefore, the answers are

a) 0.47 s

b) 1.1 m

c) 7.2 m

d) 15.9 m/s