Physics rocket problem

Question

A rocket is fired with a velocity of 9.25 x 10² m/s at 25⁰ above the horizontal. Determine how far away it lands on level ground.

Answer 1

Velocity initial = 9.25x10^2 = 925 m/s at 25 degree angle

This velocity itself is useless because we can’t plug it into any equations if it is at an angle. However, it can be used to help us find the x and y components of the initial velocity which can be plugged into kinematics equations.

SOH CAH TOA

Hypotenuse = 925

angle = 25 degrees

velocity initial x = 925cos(25) = 838.33 m/s

velocity initial y = 925sin(25) = 390.92 m/s

Let’s write our everything we know in the x direction and y direction separately:

x:

vix = 838.33 m/s

y:

viy = 390.92 m/s

velocity at top of path vtop = 0 m/s

acceleration = gravity a = -9.8 m/s^2

Here is the plan: we are looking for how far the rocket went, which is x. We only have one piece of info in the x direction, so we need to use the y direction to find something helpful for the x direction. The one variable that links the x and y directions together is time - the rocket stops moving in all directions at the same time (when it hits the ground). We will find time using the y direction and then use it in the x direction.

Possible kinematics equations:

vf = vi + at

vf^2 = vi^2 + 2ax

x = Vit + 1/2at^2

v = x/t (only when there is no acceleration)

In the y direction, we know vi, vtop, and a. We want to find t.

vf = vi + at

If we use these values to find t, we will be finding time it takes to get to the top, ttop, because we are using v top as our final velocity/stopping point.

vtop = vi + a * t top

0 = 390.92 + (-9.8)ttop subtract 390.92

-390.92 = -9.8ttop divide by -9.8

t top = 39.89 seconds

Projectile motion follows a symmetrical parabolic path, so if it took 39.89 seconds from the beginning to the top, it also took 39.89 seconds from the top back to the ground. Therefore the total time from start to finish is double t top.

t total = 2 t top = 2*39.89 = 79.78 seconds

Now in the x direction we have vix = 838.33 m/s and t = 79.78 seconds. The x direction velocity in projectile motion stays the same because the acceleration of gravity is only acting in the y direction - there is no acceleration in the x direction. That means we can use the bottom equation of our list to find the distance x.

v = x/t

838.33 = x/79.78 multiply by 79.78

x = 66881.9 m