How Do You Calculate Projectile Motion with Forces in Multiple Directions?

[ch2kb0x]

Homework Statement

A projectile fired down the y-axis with V0 = 300 ft/s, hits a target 500ft away. Assume constant force is acting on projectile at 60ft/s in negative z direction. Assume second constant force acting on projectile in negative x direction at 45ft/s.

Find angle of elevation (theta) of the launch and angle of correction (Beta) in x direction.

Find vector valued function r(t) representing path of projectile.

Find total distance traveled by projectile.

Find maximum height reached by projectile.

Find time it takes projectile to hit target.

Homework Equations

Kinematic equations.

The Attempt at a Solution

The constant force from negative z and negative x directions are completely throwing me off. Help would be appreciated, taking me through step by step. Thank you.

 

[LowlyPion]

Homework Statement

A projectile fired down the y-axis with V0 = 300 ft/s, hits a target 500ft away. Assume constant force is acting on projectile at 60ft/s in negative z direction. Assume second constant force acting on projectile in negative x direction at 45ft/s.

Find angle of elevation (theta) of the launch and angle of correction (Beta) in x direction.

Find vector valued function r(t) representing path of projectile.

Find total distance traveled by projectile.

Find maximum height reached by projectile.

Find time it takes projectile to hit target.

Homework Equations

Kinematic equations.

The Attempt at a Solution

The constant force from negative z and negative x directions are completely throwing me off. Help would be appreciated, taking me through step by step. Thank you.

You will need to modify your treatment of the Z motion with the additional velocity they provide. (And the X velocity affects X displacement.)

Note they call it a force, though it's not in the units of Force. They apparently want you to treat it as a constant velocity adjustment.

 

[nlightner]

As a scientist, it is important to approach problems systematically and use the appropriate equations and principles to solve them. Let's break down the problem and go step by step.

1. Finding the angle of elevation (theta):
We can use the kinematic equation for displacement in the y-direction to solve for the angle of elevation:
y = y0 + v0y*t + (1/2)*ay*t^2
where y0 = 0 (since the projectile is fired from the origin), v0y = V0*sin(theta), and ay = -60 ft/s^2 (since the constant force is acting in the negative z-direction).
Plugging in the values given in the problem, we get:
500 ft = 0 + (300*sin(theta))*t + (1/2)*(-60)*t^2
Simplifying, we get:
500 = 150t - 30t^2
Solving for t using the quadratic formula, we get two possible values: t = 5 s or t = -10/3 s. Since time cannot be negative, we can discard the negative solution and use t = 5 s.
Now, we can use the kinematic equation for the x-direction to solve for the angle of elevation:
x = x0 + v0x*t + (1/2)*ax*t^2
where x0 = 0, v0x = V0*cos(theta), and ax = -45 ft/s^2 (since the constant force is acting in the negative x-direction).
Plugging in the values given in the problem, we get:
500 ft = 0 + (300*cos(theta))*5 + (1/2)*(-45)*5^2
Simplifying, we get:
500 = 1500*cos(theta) - 112.5
Solving for cos(theta), we get:
cos(theta) = 0.375
Taking the inverse cosine, we get:
theta = 67.5 degrees.

2. Finding the angle of correction (Beta) in the x-direction:
Since the projectile is fired down the y-axis, there is no initial velocity in the x-direction. Therefore, the angle of correction (Beta) will be 90 degrees.

3. Finding the vector valued function r(t):
To find the path of the projectile, we can use the parametric equations for motion:
x = x0 + v0x*t + (