Answer:
[tex]a=155.750\ m/s^2[/tex]
Explanation:
It is given that,
A small rock is shoot at Dennis' neighbor's flowerpots. The rock travels a distance of 12.5 cm from the point at which he pulls it back to the point at which it leaves the slingshot.
Final velocity of the rock, v = 6.24 m/s
The initial velocity of the rock, u = 0 (as it was at rest)
Let a is the acceleration of the rock. It can be calculated using the equation of kinematics as follows :
[tex]v^2-u^2=2ad\\\\\text{Put u = 0 }\\\\v^2=2ad\\\\a=\dfrac{v^2}{2d}\\\\a=\dfrac{(6.24)^2}{2\times 12.5\times 10^{-2}}\\\\a=155.750\ m/s^2[/tex]
So, the average acceleration of the rock is [tex]155.750\ m/s^2[/tex].
