Question

A toy cannon launches a 46-g golf ball straight up into the air with a kinetic energy of 6.8 J. What must the

Answer 1

Answer : The correct option is, (C) 17 m/s

Explanation :

Formula used :

$K.E=\frac{1}{2}mv^2$

where,

K.E = kinetic energy = 6.8 J

m = mass of object = 46 g = 0.046 kg    (1 kg = 1000 g)

v = velocity

Now put all the given values in the above formula, we get:

$K.E=\frac{1}{2}mv^2$

$6.8J=\frac{1}{2}\times 0.046kg\times v^2$

$6.8kg.m^2/s^2=\frac{1}{2}\times 0.046kg\times v^2$

$v=17.19m/s\approx 17m/s$

Therefore, the ball's velocity be as it leaves the cannon is, 17 m/s