What is the Maximum Height of a Vertically Thrown Rock?

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Homework Help Overview

The discussion revolves around a physics problem involving the motion of a rock thrown vertically upwards with an initial velocity of 40.0 m/s. Participants are exploring how to determine the maximum height reached by the rock, as well as its velocity at specific time intervals.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to apply kinematic equations to find the maximum height and velocities at given times. There are questions about the timing of maximum height and the validity of calculations using different equations.

Discussion Status

Several participants are engaged in clarifying the correct approach to find the maximum height, with some suggesting the need to focus on when the velocity is zero. There is ongoing confusion regarding the calculations, with attempts to resolve discrepancies in results and interpretations of the equations.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the information they can share or the methods they can use. There is a noted lack of consensus on the correct approach and calculations, leading to varied interpretations of the problem.

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Homework Statement


Brad tosses a rock straight up at 40.0 m/s. What is the rock's velocity after 1.8 seconds? (answer 22.36 m/s)
What is the velocity of the rock from the question above after 5.2 seconds? (answer -10.96)
What is the maximum height of the rock from the question above?

Homework Equations


V=Vo+at
X=Xo+volt+1/2at^2
V^2=Vo^2+2a(X-Xo)

The Attempt at a Solution


I tried to find the maximum height for both times, but neither one was the right answer. I did the second equation (X=(40)(1.8)+1/2(-9.8)(3.24) and got 56.124 m and that wasn't right) then I tried the second equation with the second time (X=(40)(5.2)+1/2(-9.8)(27.04) and got the answer to be 75.504), but that wasn't right either. Please tell me what I am doing wrong!
*hint: the difference between the squares of the initial and final speeds is equal to twice the product of the acceleration due to gravity and the displacement.
 
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how do you know that at max height the time is 1.8s?
 
but the max height does not occur at 1.8 seconds or 5.2 seconds.
you don't know when the max height occurs,no do you care.
you only need to find the max height, so need distance traveled by rock to point when velocity is 0.
ie
u=40 v=0 (at top of motion) s=? a=-g

now use v^2=u^2+2AS
 
I tried V^2=u^2+2AS (0^2)=(40^2)+2(-9.8)s and got 0 but I don't think that's the right answer.
 
not sure how you get 0.?

you have 0=1600-19.6s

so s=0 certainly does not work
 
jojo711 said:
I tried V^2=u^2+2AS (0^2)=(40^2)+2(-9.8)s and got 0 but I don't think that's the right answer.

v^{2} = u^{2} + 2as

0 = 40^{2} + 2(-9.8)s

s = 40^{2}/[2(-9.8)]
 
Yes, but when you subtract 19.6 from 1600, you get 1580.4s and then don't you have to divide 0 by that number?
 
grzz said:
v^{2} = u^{2} + 2as

0 = 40^{2} + 2(-9.8)s

s = 40^{2}/[2(-9.8)]

I tried this and it gave me a negative number. (-81.63). It wasn't correct. I am so confused as to what to do!
 
i am not going to solve

0=1600-19.6s

for you.

surely you can see if s was negative,-19.6s would be positive so you are adding two positives to get 0 which is absurd.
 
  • #10
grzz said:
v^{2} = u^{2} + 2as

0 = 40^{2} + 2(-9.8)s

Try to be LOGICAL.

0 = 1600 - 19.6s

Now add 19.6s to both sides:

19.6s = 1600 because 19.6s + (19.6s) = 0

Now divide both sides by 19.6:

s = 1600/19.6

s = ...

Of course you do not have to show my explanation in your solution. I only did that to show you how simple math is when you try your best to be LOGICAL.
 
  • #11
i got it, thank you for your help.
 

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