Find the average velocity

  • Thread starter Mdhiggenz
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  • #1
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Homework Statement



A ball is projected upward at time = 0.0 s, from a point on a roof 10 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 98.8 m/s Consider all quantities as positive in the upward direction. The average velocity of the ball, during the first 8.79s is closest to:

Homework Equations





The Attempt at a Solution



We know that our intial velocity = 98.8m/s , and out final velocity = 0m/s

they are asking for the average velocity of the ball during the first 8.79 seconds. So I first tested out the equations which included v0,vx, t, and g.

What I tried doing at first was I tried to find the time it took for the ball to hit the ground.

so I used the formula x=xo+(v0x)t-1/2gt^2

=10+(98.8)t+1/2(-9.8)t^2

Used the quadratic formula to solve for t and got t = 0.0493s

Since the average velocity is x2-x1/t2-t1

I tried 98.8-0/0.0493-8.79 and got -11.30.

Im so confused can someone explain to me how to get the correct answer.

Which is +56 m/s
 

Answers and Replies

  • #3
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Still doesnt seem to be giving me the value I need. What am I doing incorrect?
 
  • #4
BruceW
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You don't need to find the time it takes for the ball to hit the ground. The average speed over some time is the total distance travelled in that time, divided by that time. So you don't need to do any quadratic equations, which is nice :)
 
  • #5
Curious3141
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You don't need to find the time it takes for the ball to hit the ground. The average speed over some time is the total distance travelled in that time, divided by that time. So you don't need to do any quadratic equations, which is nice :)

Just a small clarification - the question asked for average velocity, which is total displacement/time and is distinct from average speed, which is total distance over time.

Average velocity is easier to calculate here than average speed. :smile:
 
  • #6
Curious3141
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Homework Statement



A ball is projected upward at time = 0.0 s, from a point on a roof 10 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 98.8 m/s Consider all quantities as positive in the upward direction. The average velocity of the ball, during the first 8.79s is closest to:

Homework Equations





The Attempt at a Solution



We know that our intial velocity = 98.8m/s , and out final velocity = 0m/s

they are asking for the average velocity of the ball during the first 8.79 seconds. So I first tested out the equations which included v0,vx, t, and g.

What I tried doing at first was I tried to find the time it took for the ball to hit the ground.

so I used the formula x=xo+(v0x)t-1/2gt^2

=10+(98.8)t+1/2(-9.8)t^2

Used the quadratic formula to solve for t and got t = 0.0493s

Since the average velocity is x2-x1/t2-t1

I tried 98.8-0/0.0493-8.79 and got -11.30.

Im so confused can someone explain to me how to get the correct answer.

Which is +56 m/s

Start with [itex]s=ut + \frac{1}{2}at^2[/itex] where s is the displacement at time t.

Then apply Bruce's hint. Do the division by t before plugging in the numbers, it's easier.
 
  • #7
BruceW
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Just a small clarification - the question asked for average velocity, which is total displacement/time and is distinct from average speed, which is total distance over time.

Average velocity is easier to calculate here than average speed. :smile:

Ah, you're right. I was using bad terminology, I should have said average velocity, not average speed. I always forget that in 1d, speed is the absolute value of the component of velocity.
 
  • #8
327
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Ok So what I did was

S= 98.9(8.78s)-4.9*(8.78s)^2 which gave me 489.73

I then took that number and divided it by 8.78 which gave me the correct answer of 56.

What im confused about is why did we not use the height of the building, and why would we plug in the time, and then divide it by the same time.
 
  • #9
BruceW
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Well, the question is about the average velocity over some period, where the ball started at the top of the building. So the total displacement is measured from the top of the building. The ball is in projectile motion from the start, and so the change in position does not depend on the height of the building. Therefore the height of the building does not affect the total displacement.

I see it might seem counter-intuitive to plug in time for the equation for displacement, and also to divide by the same time. But this is just because in this situation, there happens to be an equation for displacement which depends on the same time period which the motion took place over. In another question, you might be told that the displacement is 52m and the time this took was 4s, in this case average velocity is 13m/s. So you see that you don't always find the average velocity by 'plugging in time, then dividing by time'. Its just that you happened to be able to do that for this specific question.
 
  • #10
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Thank you for that detailed explanation! I have been struggling alot with these types of problems but practice makes perfect!
 
  • #11
BruceW
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true, true. glad to have helped :)
 

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