Find a series of expression to describe an objects motion on a non-earth planet

Question

Note: This question is in the context of a Calculus III course.

Question:

A person throws an arbitrary object on a planet that's not earth.

The object is:

• thrown from height $h_{0}$
• thrown at an angle θ with a velocity $v_{0}$.
• The acceleration due to gravity on this planet is $g_{0}$

Find:

1. The position of the baseball as a function of time.
2. Find the "time-of-flight" of the baseball (the amount of time before it hits the ground) in terms of the angle $θ$. Use this expression to find the angle that maximizes the amount of time the ball spends in the air.
3. Find the max height of the baseball as it travels through the air as a function of θ.
4. Find the unit tangent vector for the trajectory of the baseball.
5. Find the tangential and normal components of the acceleration of the ball.

My work so far:

1. I understand that this question is looking for a series of expressions that somehow involve projectile motion

Note: I will continue to update the post as I work on the problem.

Answer 1

Have you taken a Calculus course? That's where you would see a problem like this. I would divide the position into two functions, x(t) parallel to the planet's surface and y(t) perpendicular to the planet's surface. We are told that the ball's initial height is "h". Taking the initial position on the planed to be 0, we have $x(0)= 0$ and $y(0)= h.$

Initially the ball's speed is $v_0$ at angle $\theta$ to the horizontal so $dx(0)/dt= v_0 cos(\theta)$ and $dy(0)/dt= v_0 sin(\theta).$

Finally the only force that can cause acceleration is gravity. There is no acceleration in the x-direction- $d^2x(t)/dt^2= 0$ and $d^2y(t)/dt^2= -g_0$.

Integrating $d^2x/dt^2= 0$ we get $dx/dt= constant$ which must be the same as the initial speed, $dx(0)/dt= v_0 cos(\theta)$. Integrating again, $x(t)= v_0 cos(\theta)t+ constant$. Since x(0)= 0 that constant is 0.

Integrating $d^2y/dt^2= -g_0$, $dy/dt= -g_0t+ constant$ and that constant is $v_0 sin(\theta)$. $dy/dt= -g_0t+ v_0 sin(\theta)$. Integrating again, $y(t)= -(g_0/2)t^2- v_0 sin(\theta)t+ constant$. That constant is the initial height, h: $y(t)= -(g_0/2)t^2- v_0 sin(\theta)t+ h$.

To find the "time of flight", solve the equation $y(t)= -(g_0/2)t^2- v_0 sin(\theta)t+ h= 0$ for t.

There are two ways to find the "maximum height". Since this is a parabola, opening downward, its maximum height is the y value at the vertex. Complete the square to get $y= a(t- t_0)^2+ b$. The maximum height is b and occurs at $t= t_0$. The other way is to take the derivative and set it equal to 0, $dy/dt= -g_0t- v_0sin(\theta)= 0$. That is, of course, the vertical speed.

As long as that speed is positive the ball is going up. When the speed is negative the ball is going down. It's speed is 0 at the top of the arc.

The position vector is $<x(t), y(t)>= <v_0 cos(\theta)t, -(g_0/2)t^2- v_0 sin(\theta)t+ h>.$ The tangent vector is the derivative of that to find the unit tangent vector, divide that tangent vector by its length.

The acceleration vector is always straight down, $<0, -g_0>$. Find the components of that vector parallel to and normal to that curve.