Motion under Gravity: Displacement of 273.6m in 5sec

[loku]

Homework Statement

A body is thrown vertically up from the top of a tower with a velocity of 5m/s. It reaches the ground after 5 sec. Find height of the tower.

Relevant Equations

S=ut+(1/2)at^2

u=5m/s
t=5sec
the height will be the displacement
s=ut - (1/2)gt^2
s=5*5-0.5*10* 25
s=-100m

But the answer is 273.6m

 

[berkeman]

Problem Statement: A body is thrown vertically up from the top of a tower with a velocity of 5m/s. It reaches the ground after 5 sec. Find height of the tower.
Relevant Equations: S=ut+(1/2)at^2

s=ut - (1/2)gt^2

There is an S0 term missing, no? The initial displacement is 5m up... What should that equation include to show that there is an initial displacement?

 

[loku]

I am not sure what s0 term is missing. I know only the three equations of motion.

 

[berkeman]

$$y(t) = y(0) + vy_0 * t + \frac{1}{2}a_yt^2$$

 

[loku]

we are taught:
y(t) = vy0∗t+1/2 ayt^2

 

[loku]

but still then by using the equation you gave, we don't get the correct answer.

 

[berkeman]

we are taught:
y(t) = vy0∗t+1/2 ayt^2

Please fix that typo. Thanks

 

[berkeman]

but still then by using the equation you gave, we don't get the correct answer.

Please show me your work based on the full vertical movement equation that I posted -- it should give you a good result. We can't check your work if you don't show your work. Thanks.

 

[loku]

s = 5 + 5*5 - 0.5*10*25
s = 5 + 25 -125
s = - 95 m

 

[berkeman]

A body is thrown vertically up from the top of a tower with a velocity of 5m/s. It reaches the ground after 5 sec. Find height of the tower.

$$y(t) = y(0) + vy_0 * t + \frac{1}{2}a_yt^2$$

s = 5 + 5*5 - 0.5*10*25
s = 5 + 25 -125
s = - 95 m

You seem to be substituting in variables at random. You are asked to find the initial displacement Y(t=0), so you can't plug in "5" in your initial equation for that. The initial vertical displacement in y(t) would be the unknown, and the final displacement after 5 seconds is Y(t) = 0.

Try again?

 

[SammyS]

Problem Statement: A body is thrown vertically up from the top of a tower with a velocity of 5m/s. It reaches the ground after 5 sec. Find height of the tower.
Relevant Equations: S=ut+(1/2)at^2

u=5m/s
t=5sec
the height will be the displacement
s=ut - (1/2)gt^2
s=5*5-0.5*10* 25
s=-100m

But the answer is 273.6m

Your result looks fine provided that you interpret the displacement, s, correctly.

The variable, s, is the displacement from time, t = 0 seconds, to time, t = 5 seconds. Thus the body undergoes a displacement of −100 meters in traveling from the top of the tower to the ground.

So, what is the height of the tower?

 

[loku]

shouldn't the displacement of the body be the height of the tower, since it was thrown from the height of the tower?

 

[berkeman]

shouldn't the displacement of the body be the height of the tower, since it was thrown from the height of the tower?

Sure. Which variable would that be in the equation that I posted?

Also, the initial displacement and final displacement depend on your arbitrary choice of the origin of the coordinates. Would you like to work this equation using the top of the tower or the bottom of the tower as y=0?

 

[loku]

I haven't learned about 2d motion. So simply, can we use the equation s=ut+(1/2)at^2 please?

 

[berkeman]

I haven't learned about 2d motion. So simply, can we use the equation s=ut+(1/2)at^2 please?

The equation that I posted is for 1-dimensional motion in the vertical axis with the constant acceleration of gravity acting on the object. For 2-dimensional problems, you will see an x-displacement versus time, and the general motion of the object will be an inverted parabola (like when you throw a baseball from the outfield to home plate).

 

[berkeman]

I haven't learned about 2d motion. So simply, can we use the equation s=ut+(1/2)at^2 please?

That equation does not include the initial vertical displacement, so it won't help you solve a problem like this one where the final vertical position is higher or lower than the starting vertical position. That's why you need to use the full equation that I posted. Hope that helps.

 

[loku]

Well in that case the initial displacement will be 5 m. so by using the equation, you gave
y(t) = 5 + v*5 + (1/2)(-10)t^2
for t= 5sec and v=5m/s
y(5) = 5+ 5*5 +(1/2)(-10)*25
y(5) = -95m but I know that's not correct

 

[SammyS]

shouldn't the displacement of the body be the height of the tower, since it was thrown from the height of the tower?

Is the height of the tower negative?

 

[berkeman]

Well in that case the initial displacement will be 5 m. so by using the equation, you gave
y(t) = 5 + v*5 + (1/2)(-10)t^2

No (see what I have struck out in your equation above), the initial displacement is what you are asked to find. Set y(t=5s)=0 for when it hits the ground. Initial v=5m/s, and total time when it hits the ground is t=5s. Then solve the equation for y(0).

 

[jbriggs444]

There is another approach to this problem that avoids the need for the full quadratic equation for distance moved.

1. You know the initial upward velocity and downward acceleration of gravity. From this you can compute the time to the peak of the trajectory and the resulting height gain.

2. From the peak, you know the time remaining for the fall and the downward acceleration. From this you can compute the distance fallen.

3. Subtract height gain from total fall distance to get tower height.

 

[ehild]

Also, the initial displacement and final displacement depend on your arbitrary choice of the origin of the coordinates. Would you like to work this equation using the top of the tower or the bottom of the tower as y=0?

There is initial position and final position, and the diffence between the final and initial positions is the displacement. If H is the height of the tower, the initial position is at coordinate yi=H and the final position is at yf=0. The displacement is yf-yi=-H. The OP was almost right, only he should have given the height of the tower as negative of the displacement.

 

[ehild]

Problem Statement: A body is thrown vertically up from the top of a tower with a velocity of 5m/s. It reaches the ground after 5 sec. Find height of the tower.
Relevant Equations: S=ut+(1/2)at^2

u=5m/s
t=5sec
the height will be the displacement

No, the body displaces downward while you took the upward direction positive. Just change the sign of the displacement to positive, and it is the height of the tower.

s=ut - (1/2)gt^2
s=5*5-0.5*10* 25
s=-100m

But the answer is 273.6m

No, the tower can not be that high.

 

[ehild]

There is an S0 term missing, no? The initial displacement is 5m up... What should that equation include to show that there is an initial displacement?

5 m/s is the initial velocity.

 

[ehild]

You seem to be substituting in variables at random.

You wrote in Post #2 that the "initial displacement" was 5 m.

 

[berkeman]

You wrote in Post #2 that the "initial displacement" was 5 m.

Ah, you are right. Apologies to the OP!

 

[kuruman]

But the answer is 273.6m

It cannot be. If the body were released from rest, in 5 seconds it would drop a distance
$d=\frac{1}{2}gt^2=\frac{1}{2}\times 10 \rm{(m/s^2)}\times 5^2\rm{(s^2)}=125~ \rm{m}$ which is less than 273.6 m. Since the body is not released from rest but reaches maximum height and is instantaneously at rest above the height of the building, that means that in less than 5 s it has to cover more than 125 m. No way. The answer you have is not right. You should be able to piece together the correct solution using what has already been said. I would write an equation giving the height of the body above ground at any time $t$. Then use that equation to say mathematically that "at specific time t = 5 s the body is at ground zero."

 

[loku]

so is the height 100m ??

 

[haruspex]

so is the height 100m ??

Yes.