Solve Kinematics Problem: Stone Thrown from 165m Building, 4s Travel Time

In summary, the conversation discusses a problem where a stone is thrown vertically downward from the top of a building with an initial velocity of 5.0 m/s. The question asks for the displacement during the 4th second of travel. The conversation also mentions the kinematics equation and the incorrect answer given. The correct answer is 98.4 m, but the question actually asks for the displacement during the 4th second, which is 39 m. The conversation ends with a clarification of the question and the formula for calculating displacement during a time interval.
  • #1
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Homework Statement



A student throws a stone vertically downward with an initial velocity of 5.0 m/s from the top of a 165 m building. How far does the stone travel during its 4th second of travel?

Homework Equations



d= vit + 1/2at2


The Attempt at a Solution



Vi = -5.0 m/s
t = 4.0 s
a = -9.81 m/s2
d = ?

I tried it in the kinematics equation d= vit + 1/2at2 and the answer is wrong.
 
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  • #2
Well, d=Vot+.5at^2
so d should be 98.4m. Did you get that?
 
  • #3
The answer given is 39 m.
 
  • #4
Wait, I think it was the d for the 4th second, from 3-4s. Because the first second is 0-1s, second second is for 1-2s, etc.

So
Vf=vo+at
V@3=5+9.8*3 and V@4=5+9.8*4
Vf^2=Vo^2+2ad
(V@4)^2=(V@3)^2+2*9.8*d
Find d
 
Last edited:
  • #5
d = displacement/distance.
 
  • #6
Yes, i know.
 
  • #7
Is anyone going to help solve this problem please?
 
  • #8
I just gave you the answer.
 
  • #9
As silvashadow has already pointed out, you are not reading the question carefully. It does not ask for the distance after 4 seconds, it asks for the displacement during its 4th second

Displacement during any given time interval=
(Position at final time)-(Position at initial time).

What is the final time? What is the initial?
 

Related to Solve Kinematics Problem: Stone Thrown from 165m Building, 4s Travel Time

1. What is the initial velocity of the stone thrown from the building?

The initial velocity of the stone can be found using the formula v = d/t, where v is the velocity, d is the distance, and t is the time. In this case, the distance is 165m and the time is 4s, so the initial velocity is 41.25 m/s.

2. How far does the stone travel horizontally?

The horizontal distance traveled by the stone can be calculated using the formula d = vt, where d is the distance, v is the velocity, and t is the time. In this case, the velocity is 41.25 m/s and the time is 4s, so the horizontal distance traveled is 165m.

3. What is the acceleration of the stone?

The acceleration of the stone can be found using the formula a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time. In this case, the final velocity is 0 m/s, the initial velocity is 41.25 m/s, and the time is 4s. Therefore, the acceleration is -10.3125 m/s^2, assuming the stone is thrown upwards.

4. What is the maximum height reached by the stone?

The maximum height reached by the stone can be calculated using the formula h = vi*t + (1/2)*a*t^2, where h is the height, vi is the initial velocity, t is the time, and a is the acceleration. In this case, the initial velocity is 41.25 m/s, the time is 4s, and the acceleration is -10.3125 m/s^2. Plugging these values in, the maximum height reached by the stone is 66.5m.

5. How long does it take for the stone to reach the ground?

The time it takes for the stone to reach the ground can be found using the formula t = sqrt(2h/a), where t is the time, h is the height, and a is the acceleration. In this case, the height is 165m and the acceleration is -10.3125 m/s^2. Plugging these values in, we get a time of 6.42s, which is the total time the stone takes to go up and come back down.

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