Answer : The correct option is, (C) 17 m/s
Explanation :
Formula used :
[tex]K.E=\frac{1}{2}mv^2[/tex]
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:
[tex]K.E=\frac{1}{2}mv^2[/tex]
[tex]6.8J=\frac{1}{2}\times 0.046kg\times v^2[/tex]
[tex]6.8kg.m^2/s^2=\frac{1}{2}\times 0.046kg\times v^2[/tex]
[tex]v=17.19m/s\approx 17m/s[/tex]
Therefore, the ball's velocity be as it leaves the cannon is, 17 m/s
