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

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In summary: The man on the cliff throws the ball and it returns past him 2 seconds later, with the height of the cliff being 42 meters. Neglecting air resistance and using the equation x-x0=v0(t)+.5(a)t^2, the answer is determined to be 9.8 m/s. In summary, the man threw the ball with an initial speed of 9.8 m/s.
  • #1
rdn98
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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?
 
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  • #2
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 vo = 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
 
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  • #3


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.
 
  • #4


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.
 

What is the "Ball Thrown From Cliff" experiment?

The "Ball Thrown From Cliff" experiment is a physics experiment that involves dropping a ball from a certain height on a cliff and measuring its motion and velocity as it falls.

Why is this experiment important?

This experiment is important because it helps us understand and observe the effects of gravity on objects in motion. It also allows us to calculate and analyze the motion of the ball as it falls.

What factors affect the motion of the ball in this experiment?

The motion of the ball is affected by several factors, including the height of the cliff, the initial velocity of the ball, air resistance, and the acceleration due to gravity.

What is the purpose of measuring the ball's motion and velocity?

Measuring the ball's motion and velocity allows us to analyze and understand the principles of motion and gravity. It also helps us make predictions and calculations for future experiments.

How can this experiment be applied in real life?

This experiment can be applied in real life to understand and predict the motion of objects in free fall, such as a skydiver jumping from a plane or an object falling from a building. It also has practical applications in fields such as engineering and sports.

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