Why Does the Ball Thrown Upward Return to the Man in 2 Seconds?

[rdn98]

A man on the edge of a cliff H = 42 m high throws a ball directly upward. It returns past him 2 s later. (H actually is the height of the point of release of the ball above the base of the cliff. Neglect air resistance.) (Hint - Gravity causes a downward acceleration at the rate g = 9.81 m/s2.)

There are multiple parts, but I'm only stuck on this first part.
With what initial speed did the man throw the ball?

So at first, I thought it was zero, but computer told me it was wrong so its not. For this problem, I'm taking it to be upward is positive in the y direction, and downward past the man is negative.

I'm going to use the equation x-x0=v0(t)+.5(a)t^2
where x-x0=0
t=2sec
a=9.8m/sec^2

and I get v0=-9.8. The correct answer is 9.8. So what gives?

 

[frankR]

The a in your equation (the acceleration of gravity at the Earth's surface) should be negative, if you want the direction down towards the center of the Earth to be negative. This should fix the sign to be v_o = 9.8 m/s.

Note: When doing problems draw a picture and label it with a frame, i.e. up/down (y-axis) and left/right (x-axis). Even when working with other areas of physics, it's critically important to place a frame of reference.

Frank

 

[M98Ranger]

The issue here is that you have used the acceleration due to gravity as a positive value in your equation. Since the ball is thrown upward, the initial velocity (v0) should also be positive. This means that the acceleration due to gravity should be negative in your equation, since it is acting in the opposite direction of the initial velocity.

So, the correct equation to use in this situation would be x-x0=v0(t)-.5(g)t^2, where g is now negative since it is acting in the downward direction. Plugging in the given values, we get:

0 = v0(2) - .5(9.81)(2)^2

Solving for v0, we get v0 = 9.81 m/s, which is the correct answer. Remember to pay attention to the direction of the variables in your equations, as it can affect the outcome of your calculations.

 

[M98Ranger]

It looks like you are on the right track, but you made a small mistake in your calculation. When using the equation x-x0=v0(t)+.5(a)t^2, you need to make sure that all of your units are consistent. In this case, you are using meters for distance and seconds for time, but you are using meters per second squared for acceleration. This is why you are getting a negative answer for v0.

To fix this, you need to convert the acceleration from meters per second squared to meters per second per second. This will give you the correct units for velocity.

So, using the equation v0=(x-x0-.5(a)t^2)/t, we get v0=(0-42-.5(9.81)(2)^2)/2, which gives us v0=9.81 m/s.

In summary, make sure to always check your units when using equations to solve problems. In this case, the correct units for acceleration should be meters per second per second, not meters per second squared.