A person stands on the ground with a slingshot and tries to hit a high limb on a nearby tree. The limb is 18.3 feet above the ground. What is the minimum initial velocity in meters/second of a rock leaving the slingshot at 6 feet above the ground that would be able to reach the limb

Respuesta :

The minimum initial velocity of the rock is V = 8.56 m/sec.

What is Conservation of Energy?

According to the rule of conservation of energy, energy can only be transformed from one form of energy to another and cannot be created or destroyed. This indicates that unless energy is added from the outside, a system always has the same quantity of energy.

Calculate the rock's minimal initial velocity in meters/second as it exits the slingshot at a distance of 6 feet [tex]$\left(h_{2}=6\right.ft.)[/tex] above the ground.

Using the provided information, the net height of the limb is calculated as shown below.

[tex]$h &=h_{1}-h_{2}$[/tex]

h = 18.3 - 6

h = 12.3

Now, using the conservation of energy the minimum initial velocity of the rock is computed as shown below,

Kinetic energy of the rock = Potential energy the rock

[tex]$\begin{aligned}\frac{1}{2} m V^{2} &=m g h \\\frac{1}{2} V^{2} &=g h \\\frac{1}{2} V^{2} &=\left(32.1 \mathrm{ft} / \mathrm{s}^{2}\right)(12.3 \text { feet }) \\V &=\sqrt{2\left(32.1 \mathrm{ft} / \mathrm{s}^{2}\right)(12.3 \mathrm{feet})}\end{aligned}$[/tex]

V = 28.10 [tex]\frac{ft}{s}[/tex].

The minimal initial velocity in meter/second is 8.56.

Therefore, the minimum initial velocity of the rock is 8.56 m/sec

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