Why is my calculation for the height of the building off?

  • Thread starter Toranc3
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In summary, the problem involves two balls being thrown from the edge of a building - one is thrown straight up and the other is dropped 1 second later. The goal is to find the height of the building for both balls to reach the ground at the same time, assuming no air resistance. Using the equations y=yo+vo*t+1/2*a*t^(2) and t2=t1-1seconds, we can set the two final equations equal to each other and solve for t1, which is approximately equal to 1.2874 seconds. However, there may be some slight variations in the answer due to rounding errors or the use of different values for acceleration due to gravity (e.g. 9.8
  • #1
Toranc3
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Homework Statement



A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1 second later. You may ignore air resistance

What must the height of the building be for both balls to reach the ground at the same time if vo is 6.0m/s?

Homework Equations



y=yo+vo*t+1/2*a*t^(2)

t2=t1-1seconds


The Attempt at a Solution



I chose positive going down

ball 2:

y=yo+vo*t +1/2a*t^(2)

y=4.905m/s^(2)*t2^(2)
y=4.905m/s^(2)*(t1-1s)^(2)

y=4.905m/s^(2)t1^(2)-9.81m/s*t1+4.905m

ball 1

y=yo+vo*t +1/2a*t^(2)

y=-6m/s*t1 + 4.90m/s^(2) *t1^(2)

Set the final equations equal to each other:
y=y

4.905m/s^(2)t1^(2)-9.81m/s*t1+4.905m=-6m/s*t1 + 4.90m/s^(2) *t1^(2)

4.905m=3.81m/s*t1

t1=1.2874 seconds

plug t1 back into equation for ball 1

I get 0.4051 m but the answer is 0.411 m

What did I do wrong?
 
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  • #2
Toranc3 said:

Homework Statement



A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1 second later. You may ignore air resistance

What must the height of the building be for both balls to reach the ground at the same time if vo is 6.0m/s?

Homework Equations



y=yo+vo*t+1/2*a*t^(2)

t2=t1-1seconds


The Attempt at a Solution



I chose positive going down

ball 2:

y=yo+vo*t +1/2a*t^(2)

y=4.905m/s^(2)*t2^(2)
y=4.905m/s^(2)*(t1-1s)^(2)

y=4.905m/s^(2)t1^(2)-9.81m/s*t1+4.905m

ball 1

y=yo+vo*t +1/2a*t^(2)

y=-6m/s*t1 + 4.90m/s^(2) *t1^(2)

Set the final equations equal to each other:
y=y

4.905m/s^(2)t1^(2)-9.81m/s*t1+4.905m=-6m/s*t1 + 4.90m/s^(2) *t1^(2)

4.905m=3.81m/s*t1

t1=1.2874 seconds

plug t1 back into equation for ball 1

I get 0.4051 m but the answer is 0.411 m

What did I do wrong?

I have not checked your calculations closely but a couple of things you may look for.

Your answer is very close, I wonder if there is:
(a) some rounding errors?
(b) the setter used g = 9.8 rather than 9.81?
(c) the setter used g = 10 rather than 9.81?
 
  • #3
PeterO said:
I have not checked your calculations closely but a couple of things you may look for.

Your answer is very close, I wonder if there is:
(a) some rounding errors?
(b) the setter used g = 9.8 rather than 9.81?
(c) the setter used g = 10 rather than 9.81?

Yeah your right. Thanks buddy!
 
  • #4
You say in your post "I chose positive going down". If you do that, and the ball were thrown up, [itex]v_0[/itex] must be negative.
 
  • #5
HallsofIvy said:
You say in your post "I chose positive going down". If you do that, and the ball were thrown up, [itex]v_0[/itex] must be negative.

I cat and pasted from original

y=-6m/s*t1 + 4.90m/s^(2) *t1^(2)

he had -6 ??
 

Related to Why is my calculation for the height of the building off?

1. What is the initial velocity of a ball thrown straight up?

The initial velocity of a ball thrown straight up is the velocity at which it leaves the hand of the person throwing it. This can vary depending on factors such as the strength and technique of the person throwing the ball.

2. How long will a ball take to reach its maximum height when thrown straight up?

The time it takes for a ball to reach its maximum height when thrown straight up can be calculated using the formula t = √(2h/g), where h is the initial height and g is the acceleration due to gravity (9.8 m/s² on Earth). This means that the time taken will depend on the initial height of the ball.

3. What is the acceleration of a ball thrown straight up?

The acceleration of a ball thrown straight up is equal to the acceleration due to gravity, which is 9.8 m/s² on Earth. This means that the ball will experience a constant acceleration of -9.8 m/s² as it moves upwards and then downwards.

4. How high will a ball go when thrown straight up?

The height that a ball will reach when thrown straight up can be calculated using the formula h = (v₀²)/2g, where v₀ is the initial velocity of the ball. This means that the height will depend on the initial velocity of the ball, which can vary depending on the person throwing it.

5. What happens to the velocity of a ball as it moves up and then down when thrown straight up?

The velocity of a ball when thrown straight up will decrease as it moves upwards due to the acceleration of gravity. At the highest point, the velocity will be 0 m/s. As the ball starts to fall back down, the velocity will increase due to the acceleration of gravity, reaching the same velocity as it had when thrown from the start. This means that the velocity of a ball thrown straight up will change from positive to negative and back to positive again.

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