- #1

Bashyboy

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## Homework Statement

A cannon shoots a ball at an angle θ above the horizontal ground. Neglecting air resistance use newton's second law to find the ball's position as a function of time. (Use the axes with x measured horizontally and y vertically.) Let r(t) denote the balls distance from the cannon. What is the largest possible value of θ if r(t) is to increase throughout the ball's flight? (Hint: using your solution to part (a) you can write down r^2 as x^2 + y^2, and then find the condition that r^2 is always increasing.)

## Homework Equations

## The Attempt at a Solution

Suppose that the magnitude of the initial velocity is [itex]v_0[/itex]. Since we are neglecting air-resistance, no force opposes the motion directed parallel to the x-axis.

[itex]v_x = v_0 \cos \theta[/itex] (velocity does not change with time)

Integrating, [itex]x(t) = v_0 (\cos)t \theta + x_0[/itex]

The only force acting on the object is the gravitational force.

[itex]F_g = ma[/itex]

rewriting

[itex]a = F_g/m[/itex]

Integrating twice, [itex]y(t) = - \frac{1}{2}gt^2 + v_{y0}t + y_0[/itex], where g = F_g/m and v_yo = v_0 sinθ.

y_0 and x_0 are the initial positions of the object. Define the origin of the coordinate system to coincide with the initial point. Hence, (x_0,y_0) = (0,0)

Now, the vector position function [itex]\vec{r(t}[/itex] can be written as [itex]\vec{r(t)}= (v_0 cos \theta t) \hat{i} + (gt^2 + v_0 sin (\theta) t[/itex]

The magnitude of this is:

[itex]| \vec{r(t)} | = \sqrt{v^2_0 cos^2 \theta - \frac{1}{4} g^2 t^4 + gt^3 v_0 sin \theta + v_0^2 sin^2 (\theta) t^2}[/itex]

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Does this appear correct? Also, when it says that you can write r^2 as x^2 + y^2, are they asking me to convert my position into polar coordinates?

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