Question

A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find the ball's position as a function of time. (Use axes with x measured horizontally and y vertically.) (b) Let r(t) denote the ball's distance from the cannon. What is the largest possible value of θ if r(t) is to increase throughout the ball's flight?

Answer 1

Answer:

a)  x = v₀ₓ t ,  y = $v_{oy}$ t - ½ g t²

Explanation:

This is a projectile launch problem, let's use Newton's second law on each axis

X axis

       F = m a

Since there is no acceleration on the x axis, the force on this axis is zero

Y Axis  

       -W = m $a_{y}$

       -m g = m $a_{y}$  

       $a_{y}$  = -g

In this axis the acceleration is the acceleration of gravity

Now we can use science to find the position of the body on each axis

X axis

       x = v₀ₓ t + ½ a  t²

As the acceleration on this axis is zero

      x = v₀ₓ t

Y Axis

       y = $v_{oy}$ t + ½ $a_{y}$  t²

The acceleration on this axis is –g

        y = $v_{oy}$ t - ½ g t²

B) to find the maximum value of distance r

        r =√ x² + y²

        r = √( v₀ₓ² t² + ($v_{oy}$ t + ½ g t²)²

We can find the maximum value of r using time respect derivatives

      dr / dt = 0

      0 = ½ 1/√( v₀ₓ² t² + ($v_{oy}$ t + ½ g t²)²    (v₀ₓ² 2t + 2 ($v_{oy}$ t - ½ g t²)($v_{oy}$ - ½ g2t)

We simplify this expression

         0 = v₀ₓ² 2t + 2 ($v_{oy}$ t - ½ g t²)  ($v_{oy}$ - ½ g2t)

       -v₀ₓ² t =( $v_{oy}$ t - ½ g t²)  ($v_{oy}$ - ½ g2t)  

      -v₀ₓ² t = $v_{oy}$² t -3/2 gt² $v_{oy}$ + 1/2 g² t³  

       ½ g² t²- 3/2 g $v_{oy}$ t  = $v_{oy}$² + v₀ₓ²

Let's use trigonometry to find go and vox

         sin θ = $v_{oy}$ / v₀

         cos θ = vox / v₀

         $v_{oy}$ = v₀ sin θ

        v₀ₓ = v₀ cos θ

We replace

         ½ g² t² -3/2 g $v_{oy}$ t = v₀ (sin² θ +  cos² θ)

          g t² - 3v₀ sin θ t = 2 v₀/g

 

The time is maximum for the angle is zero