# Calculate the displacement? (Projectile motion problem)

• TarPaul91
In summary: You can do this by solving the following equation in polar coordinates: $$\rho(t) = \textbf{r}_0(\textbf{P}(t))$$where r0 is the initial radius. In summary, the 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. It has an initial speed of 560.16 m/s and a range of 2778 m.
TarPaul91

## 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

## The Attempt at a Solution

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

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

Regarding #4, let P(t) be the location of the projectile at time t. You want to find |P(91) - P(0)|.

## What is the definition of displacement in projectile motion?

Displacement in projectile motion is the change in position of an object from its initial position to its final position. It is a vector quantity that includes both magnitude and direction.

## How do you calculate displacement in projectile motion?

Displacement in projectile motion can be calculated using the equation: Δx = v0t + (1/2)at2, where Δx is displacement, v0 is initial velocity, t is time, and a is acceleration.

## What is the difference between displacement and distance in projectile motion?

Distance in projectile motion is the total length of the path an object travels, while displacement is the shortest distance between the initial and final positions of the object. Distance is a scalar quantity, while displacement is a vector quantity.

## How does the angle of projection affect displacement in projectile motion?

The angle of projection affects the horizontal and vertical components of displacement. A larger angle will result in a greater horizontal displacement, while a smaller angle will result in a greater vertical displacement.

## Can you have negative displacement in projectile motion?

Yes, displacement can be negative in projectile motion. This indicates that the object has moved in the opposite direction of the initial velocity. In other words, the final position is behind the initial position.

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