Calculate the displacement? (Projectile motion problem)

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A projectile launched at a 60° angle lands after 99 seconds, with initial speed calculated at 560.16 m/s, a maximum altitude of 12007 m, and a range of 2778 m. The main challenge lies in calculating the displacement at 91 seconds, which requires using the SUVAT equations under constant acceleration. The horizontal displacement remains constant while the vertical displacement can be derived using the initial vertical speed and gravitational acceleration. To find the overall displacement, the position vectors at 91 seconds and the launch point must be considered. The discussion emphasizes the need for a clear understanding of vector components to solve for displacement effectively.
TarPaul91
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Homework Statement


A projectile is launched at an angle of 60° from the horizontal and lands 99 s later at the same height from which it was launched.
1) What is the initial speed of the projectile (in m/s)?
I already figured this out to be 560.16 m/s
2) What is the maximum altitude (in m)?
- I already figured this out to be 12007 m
3) What is the range (in m)?
- I already figured this out to be 2778 m
4) Calculate the displacement (in m) from the point of launch to the position on its trajectory at 91 s. (Express your answer in vector form. Assume the projectile initially travels in the +x and +y-directions, where the +x-direction is horizontal and the +y-direction is straight up.)
^^^^This is where I got lost

Homework Equations



The Attempt at a Solution


I was able to figure out parts 1-3, but I'm not sure how to calculate the displacement.
 
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TarPaul91 said:

Homework Statement


4) Calculate the displacement (in m) from the point of launch to the position on its trajectory at 91 s. (Express your answer in vector form. Assume the projectile initially travels in the +x and +y-directions, where the +x-direction is horizontal and the +y-direction is straight up.)

if we are assuming that acceleration is constant -g, then we can use SUVAT equations to solve this:
$$ \textbf{s} = \textbf{u}t + \frac{1}{2}\textbf{a}t^2$$
becomes
$$ \textbf{s} = \textbf{u}t - \frac{1}{2}\textbf{g}t^2$$

You know what u and g vectors are and you can use components to solve for s... I will leave it there with you. If you need more help, please ask and I am happy to provide more hints.
 
Hi
I’ll use two equations to derive something
We know that horizontal direction has no force this velocity remains constant = ucos(theta)
This displacement along x direction is
X=ucos(theta)xtime
Also Y=usin(theta).t -1/2g.t^2
Substitute for t using the first equation to get an equation independent of time
That is known as equation of trajectory.
This specific thing may help you out solving for displacement at 91seconds
 
@TarPaul91, your answer for #3 is wrong.

Regarding #4, let P(t) be the location of the projectile at time t. You want to find |P(91) - P(0)|.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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