A person stands on the ground with a slingshot and tries to hit a high limb on a nearby tree. The limb is 23.8 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? Note: Convert units as necessary.

Respuesta :

Answer:

The minimum initial velocity is 10.31 m/s.

Explanation:

Given that,

Length = 23.8 feet

We need to calculate the height of the limb from the sling

[tex]h=23.8-6=17.8\ feet[/tex]

Using conservation of energy theorem

[tex]M.E_{i}=M.E_{f}[/tex]

[tex]P.E_{i}+K.E_{i}=P.E_{f}+K.E_{f}[/tex]

[tex]0+\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=mgh+0[/tex]

[tex]\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=mgh[/tex]

At position of limb, final velocity is zero.

[tex]v_{f}=0[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}v_{i}^2=gh[/tex]

[tex]v_{i}=\sqrt{2gh}[/tex]

Put the value into the formula

[tex]v_{i}=\sqrt{2\times9.8\times5.42544}[/tex]

[tex]v_{i}=10.31\ m/s[/tex]

Hence, The minimum initial velocity is 10.31 m/s.