A ball is thrown horizontally from a height of 24 m and hits the ground with a speed that is four times its initial speed. What is the initial speed?

Question

austinl · Answer

@Jemurray3 , would you mind terribly helping me out with this problem?

frostbite · Answer

Can't we use kinematics on this one?

austinl · Answer

Probably, I am drawing a big fat blank on this. It seems like it should be really simple...

frostbite · Answer

And when I come to think about it... I don't think so... kinematics usually require a time unit.

austinl · Answer

I was just looking at a list of equations, and they all require time or horizontal distance.

anonymous · Answer

You can find time by recognizing how long it will take for that ball to hit the ground.

austinl · Answer

Yeah, 2.45 seconds.

anonymous · Answer

If you know conservation of energy, you could use that as well...and it would be considerably easier.

austinl · Answer

I think I know it, I will have to look it up.

austinl · Answer

KE+PE=constant

austinl · Answer

Excuse me, t=2.21 seconds I believe.

austinl · Answer

I got 5.60 m/s. Is this correct?

anonymous · Answer

That means that
$$ mgh + \frac{1}{2} mv_i^2 = \frac{1}{2} m v_f^2$$
or, dividing by mass and multiplying by 2,
$$ v_f^2 - v_i^2 = 2gh $$
If you're given that vf = 4 vi, then that gives
$$v_f^2-v_i^2 = 15 v_i^2 = 2 gh \implies v_i = \sqrt{\frac{2gh}{15}}  = 5.6 \space m/s$$

austinl · Answer

That is how I did it.