What is the speed of the ball immediately before it strikes the ground?

In summary, the conversation revolves around finding the speed of a ball as it strikes the ground after being shot vertically upwards with an initial speed of 10m/s and a starting height of 5m. The formula Vf= Vi+g.T is discussed, along with the use of acceleration due to gravity (-9.8m/s2). The concept of energy conservation is also introduced as a simpler method to solve the problem. Both methods are suggested for a better understanding of the topic.
  • #1
rmackay
3
0
Hi, I am new here, and am not expecting any direct answers (I read the sticky) but I am pretty confused about this one question:
A ball is shot vertically upwards at 10m/s from a height of 5m above the ground, If friction can be ignored, what is the speed of the ball immediately before it strikes the ground?

I need to know if the formula I am using is correct:
Vf= Vi+g.T
would g (9.8m/s2) be -9.8m/s2 for this question because it uses deceleration?
thanks alot!
 
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  • #2
Welcome to PF!
It is certainly A correct formula, and if you choose Vi to be positive (that is, upwards direction positive), you should definitely use the negative -9.8.

However, as yet, T is also an unknown, not only Vf!
So, (it seems..) you need yet another equation in order to close your system.
Do you have any idea as to what that might be?
 
  • #3
Your formula is correct; the acceleration due to gravity is constant, so the total change in acceleration from launch to landing is gt.

However, you're going about this problem the hard way, since you'd need to find t first to use your formula, and finding t in this case could be a bit complicated.

How about using an energy argument instead? The ball begins with a specific amount of kinetic energy, which it trades for altitude as it rises to the top of its trajectory. It then begins to fall, and trades that altitude back into kinetic energy.

Since there is no friction, no energy is lost. The ball is going the same speed when it was launched upwards from 5 m altitude as when falling downward past 5 m altitude. In other words, it's going 10 m/s downwards as it passes 5 m altitude.

Now, the problem is significantly simplified! All you need to do is determine how much speed the ball would gain in falling the additional 5 meters to the ground, then add it to 10 m/s it had at 5 m altitude.

This doesn't require time at all; all you need to do is figure out how much gravitational potential energy the ball lost in moving downwards 5 meters ([itex]= mg\Delta h[/itex]). All of that gravitational potential energy is converted into kinetic energy (so [itex]mg\Delta h = (1/2) m(\Delta v)^2[/itex]). Can you solve that equation for [itex]\Delta v[/itex]?

Hint: This is an extremely common sort of thing to be asked in a high-school physics class -- how much speed does a ball gain in falling x meters. You might want to remember the very simple formula you'll find.

- Warren
 
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  • #4
I fully concur with chroot here; an energy argument is definitely the simplest (I gave an oblique hint to that).

However, if you are at the beginning of learning physics, it can be quite helpful to verify that energy arguments and a plodding solution of Newton's second law directly gives the same answers (which must "obviously" be true)
 
  • #5
t = (2 ðy/g)1/2
to get the time?
 
  • #6
Finding the time (if you decide to go that route) involves solving the position equation:

[tex]y(t) = y_0 + v_0 t + \frac{1}{2} g t^2[/tex]

where [itex]y_0 = 5[/itex] and [itex]v_0 = 10[/itex].

- Warren
 
  • #7
I think I am going to use both methods so I am sure that I am correct, it will take a while, but at least I will know how to answer this type of question in the future! I will be sure to remember what a energy argument is!
thanks again!
 
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  • #8
rmackay:

Feel free to post your intermediate results for us, so we can make sure you stay on the right track. :smile:

- Warren
 

1. What is the speed of the ball immediately before it strikes the ground?

The speed of the ball immediately before it strikes the ground is dependent on various factors such as the initial speed of the ball, the angle at which it was thrown, and the effects of air resistance. Therefore, it is not possible to determine the exact speed without knowing these variables.

2. How does the speed of the ball change as it approaches the ground?

As the ball approaches the ground, its speed will increase due to the force of gravity. This increase in speed will continue until the ball reaches its maximum velocity, also known as terminal velocity, where the force of air resistance balances out the force of gravity.

3. Can the speed of the ball be calculated using the laws of motion?

Yes, the speed of the ball before it strikes the ground can be calculated using Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. However, as mentioned before, the calculation will also depend on other variables such as air resistance.

4. How does the surface on which the ball lands affect its speed?

The surface on which the ball lands can have an impact on its speed. If the surface is soft and cushiony, the ball's speed will decrease as the surface absorbs some of its kinetic energy. On the other hand, if the surface is hard and rigid, the ball's speed may increase or remain the same as it bounces off the surface.

5. Is there a way to measure the speed of the ball as it strikes the ground?

Yes, the speed of the ball as it strikes the ground can be measured using various methods such as high-speed cameras, radar guns, and motion sensors. These devices can accurately measure the speed of the ball and provide valuable data for further analysis and research.

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