Kinetic Energy and Time Problem

[jtim36]

Homework Statement

A ball with mass m projected horizontally o the end of a table with an initial kinetic energy K. At a time t after it leaves the end of the table it has kinetic energy 3K. What is t? Neglect air resistance.
(answer: (2/g)(√K/m)

Homework Equations

E = 1/2 mv^2

The Attempt at a Solution

I understand what the problem is asking, but I don't get how to relate time and kinetic energy

 

[haruspex]

What equations do you know for the motion of a projectile under gravity?

 

[jtim36]

x = vo(t) + 1/2at^2
v^2 = v0^2 + 2ah

Okay, so does that mean the final velocity was tripled then? and I apply the above equations?

 

[Dick]

x = vo(t) + 1/2at^2
v^2 = v0^2 + 2ah

Okay, so does that mean the final velocity was tripled then? and I apply the above equations?

No. Not tripled. Think conservation of energy. You have to gain energy 2K by losing potential energy mgh. Put time into it by figuring how long it takes to fall a distance h with zero initial vertical velocity.

 

[Nickie]

in this situation.
The solution to this problem involves using the equation for kinetic energy, E = 1/2 mv^2, and setting it equal to the given values of K and 3K at times t=0 and t=t, respectively. By solving for v in both equations and setting them equal to each other, you can find the relationship between v and t. Then, using the fact that the ball is projected horizontally, you can use the equation for horizontal displacement, x = vt, to solve for t. The final solution should be t = (2/g)(√K/m), where g is the acceleration due to gravity. This solution assumes that the ball is not affected by air resistance, which is stated in the problem.