What is the Maximum Height of a Vertically Thrown Rock?

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In summary, Brad throws a rock up at 40.0 m/s and it has a velocity of 22.36 m/s after 1.8 seconds. The rock has a velocity of -10.96 after 5.2 seconds. The rock's maximum height is 75.504 m.
  • #1
jojo711
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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+Vot+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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  • #2
how do you know that at max height the time is 1.8s?
 
  • #3
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
 
  • #4
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.
 
  • #5
not sure how you get 0.?

you have 0=1600-19.6s

so s=0 certainly does not work
 
  • #6
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[itex]^{2}[/itex] = u[itex]^{2}[/itex] + 2as

0 = 40[itex]^{2}[/itex] + 2(-9.8)s

s = 40[itex]^{2}[/itex]/[2(-9.8)]
 
  • #7
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?
 
  • #8
grzz said:
v[itex]^{2}[/itex] = u[itex]^{2}[/itex] + 2as

0 = 40[itex]^{2}[/itex] + 2(-9.8)s

s = 40[itex]^{2}[/itex]/[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!
 
  • #9
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[itex]^{2}[/itex] = u[itex]^{2}[/itex] + 2as

0 = 40[itex]^{2}[/itex] + 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.
 

1. What factors affect the trajectory of a vertically thrown rock?

The trajectory of a vertically thrown rock is affected by factors such as the initial velocity, the angle of release, air resistance, and the force of gravity.

2. How does air resistance impact the motion of a vertically thrown rock?

Air resistance, also known as drag, opposes the motion of a vertically thrown rock and causes it to slow down. This can impact the trajectory and distance traveled by the rock.

3. Can the angle of release affect the distance a vertically thrown rock travels?

Yes, the angle of release can greatly impact the distance a vertically thrown rock travels. A higher angle will result in a shorter distance, while a lower angle will result in a longer distance.

4. Is the force of gravity constant for a vertically thrown rock?

Yes, the force of gravity remains constant for a vertically thrown rock. However, its effect can be influenced by other factors such as air resistance and the angle of release.

5. How does the initial velocity impact the motion and trajectory of a vertically thrown rock?

The initial velocity, or the speed at which the rock is thrown, directly affects its motion and trajectory. A higher initial velocity will result in a longer distance traveled, while a lower initial velocity will result in a shorter distance traveled.

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