Kevin S. Huang

2010: Problem 12

Topic: Energy
Concepts: Kinetic energy, Projectile motion

Solution:

By definition of kinetic energy,

K=\frac{1}{2}mv_x^2

where $v_x$ is the initial horizontal velocity of the ball.

Since the ball is undergoing projectile motion, its horizontal velocity stays constant. After time $t$ , its vertical velocity is

v_y=gt

We are given the kinetic energy at this point is $3K$ so we have

3K=\frac{1}{2}mv^2=\frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2=K+\frac{1}{2}m(gt)^2

4K=mg^2t^2

t=\frac{2}{g}\sqrt{\frac{K}{m}}

so the answer is B.