Need help on direction of acceleration

In summary, Andrew measured the time it took for a stone, thrown by David to return to the height of the bridge to be .5 seconds. He found that the stone would travel a total of 9.806 meters before it reached the water.
  • #1
paperdoll
69
0

Homework Statement


The bridge is 9.5 m above the water and using his watch, Andrew measured the time it took for a stone, thrown by David to return to the height of the bridge to be .5 seconds.

a. what is the vertical velocity with which the stone left David's hand?
b. how high did the stone go over the water?
c. what was the magnutude of the stones velocity in the horizontal direction if it left david's hand at an angle of 38 degrees?
d. what distance from the base of the bridge did the stone travel?

Homework Equations


s=ut+1/2at^2

The Attempt at a Solution


The thing I am confused about is whether to put the acceleration negative or positive.
I let Vs=0
and then
0=ut+1/2at^2
0=0.5u-9.8*1/2*(0.5)^2
u=2.45 m s^-1

for part b) I did
At max height v=0
0=2.45^2+2*-9.8*s
s=0.306
so max height is 9.5+0.306?= 9.806 m

c)
tan 38=2.45/uH
uH=3.1358 ms^-1

d) sH=uH*t
3.1358*t
v^2=2.45^2-2*9.8*-9.5
...
and then I get a negative answer that cannot be square rooted :\

Is what I'm doing right? I'm not sure on the direction of acceleration :\
 
Last edited:
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  • #2
Hi! :smile:

d) sH=uH*t
3.1358*t

Shouldn't you just fill in t=0.5 sec here?

v^2=2.45^2-2*9.8*-9.5
What are you calculating here?
 
  • #3
I like Serena said:
Hi! :smile:
Shouldn't you just fill in t=0.5 sec here?What are you calculating here?

Hi :)
I thought t=0.5 was the time it takes for the ball to go up, and then reach the same height as the bridge. So with the Sh=Uh*t formula shouldn't the "t" be total time it takes for the ball to fall down again?

On that thing, I am trying to find the final velocity so I can find the total "t"
 
  • #4
paperdoll said:
Hi :)
I thought t=0.5 was the time it takes for the ball to go up, and then reach the same height as the bridge. So with the Sh=Uh*t formula shouldn't the "t" be total time it takes for the ball to fall down again?

On that thing, I am trying to find the final velocity so I can find the total "t"

Well, you're problem statement isn't very clear, but I was thinking that "thrown by David to return to the height of the bridge" means that David threw the stone from the height of the bridge, and that the stone lands at the height of the bridge.
At least that is how you calculated (a), (b), and (c) if I'm not mistaken.
In particular it would mean that 0.5 seconds is the total time to go up and go down again.

Or am I misunderstanding the problem statement?
 
  • #5
I like Serena said:
Well, you're problem statement isn't very clear, but I was thinking that "thrown by David to return to the height of the bridge" means that David threw the stone from the height of the bridge, and that the stone lands at the height of the bridge.
At least that is how you calculated (a), (b), and (c) if I'm not mistaken.
In particular it would mean that 0.5 seconds is the total time to go up and go down again.

Or am I misunderstanding the problem statement?

I'm a bit confused because the ball needs to go down to the water so it will go up, then go back to the same height as the bridge, and then go back down 9.5 m to the water...so total time shouldn't be 0.5 seconds
 
  • #6
paperdoll said:
I'm a bit confused because the ball needs to go down to the water so it will go up, then go back to the same height as the bridge, and then go back down 9.5 m to the water...so total time shouldn't be 0.5 seconds

Ah, okay.
Btw, is it a stone or a ball? ;)

So the ball would go up for 0.25 seconds, down again for 0.25 seconds at which time it passes the bridge, and then further down until the water, yes?

So what you would like to know is the total time, so you can use that in your formula sH=uH*t, right?

How much time does it take the ball to drop from its highest point to the water?
 
  • #7
I like Serena said:
Ah, okay.
Btw, is it a stone or a ball? ;)

So the ball would go up for 0.25 seconds, down again for 0.25 seconds at which time it passes the bridge, and then further down until the water, yes?

So what you would like to know is the total time, so you can use that in your formula sH=uH*t, right?

How much time does it take the ball to drop from its highest point to the water?

Oh! so I should find the time to drop from the highest point and then add 0.25?
 
  • #8
paperdoll said:
Oh! so I should find the time to drop from the highest point and then add 0.25?

Yep. That's the easiest method I can think of.

http://www.circuitgallery.com/media/catalog/product/cache/1/small_image/9df78eab33525d08d6e5fb8d27136e95/M/a/MacMahon_abc005_250.jpg [Broken]
 
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1. What is acceleration?

Acceleration is a physical quantity that measures the rate of change of an object's velocity over time. It is a vector quantity, meaning it has both magnitude and direction.

2. How is acceleration calculated?

Acceleration can be calculated by dividing the change in velocity by the change in time. This is represented by the equation a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What is the difference between acceleration and velocity?

Velocity is a measure of an object's speed and direction, while acceleration is a measure of how quickly an object's velocity is changing. Velocity is a vector quantity, while acceleration is a vector quantity.

4. How does direction affect acceleration?

The direction of acceleration is determined by the direction of the change in velocity. If an object's velocity is increasing, the acceleration will be in the same direction as the velocity. If the velocity is decreasing, the acceleration will be in the opposite direction.

5. How does mass affect acceleration?

According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force applied to the object and inversely proportional to its mass. This means that a larger mass will require a greater force to achieve the same acceleration as a smaller mass.

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