# Minimum speed using calculus

I've been trying to answer this question , specifically part c)

I've done part a) and b) where $U = \frac{28\sqrt{15}}{15}$ and $\theta = 53.3$ degrees . For the last part I need to work out the minimum value of v. As soon as I saw the word 'minimum' , I thought I had to use calculus to work it out. I've seen the worked answer to this but calculus is not used; he says the minimum value of v is when the vertical velocity is 0 which I do understand. My working out for this does not yield the correct answer to which I do not understand why hence I can only conclude something is wrong with what I'm doing.

My method was to find the value of $\theta$ for which $v$ is a minimum. I did this by forming an equation in $v$ from the standard equation of motion $v = u + at$ where in my case:

$v = usin(\theta) - gt$

Then

$\frac{dv}{d\theta} = ucos(\theta) = 0$

Then solve that which gives $\theta$ as $\frac{\pi}{2}$

Then I plugged that into my original equation to get a value of v for which I thought was minumum but I ended up with something like $-12.37$ which is incorrect.

The correct answer IIRC is 4.32 m/s.

Can someone show me how to derive the correct answer using calculus or show me where I went wrong ? I suspect it's my equation of v.

The velocity of the ball in the horizontal direction is constant until it hits the ground. The velocity of the ball in the vertical direction varies, and becomes $0$ when the ball reaches its maximum height.

The speed at time $t$ is $\sqrt{v_h^2(t)+v_v^2(t)}$, where $v_h(t)$ is the horizontal component of velocity at time $t$, and $v_v(t)$ is the vertical component. Since $v_h(t)$ is constant, the minimum speed is reached when $v_v(t)=0$.

Thus the minimum speed is the speed in the horizontal direction. This is $u\cos\theta$.

If you want to use calculus you have to calculate the speed $s$, with respect to time. A bit of trig shows that $s(t) = \sqrt{\text{verticalspeed}(t)^2 + \text{horizontalspeed}(t)^2}$ the horizontal speed is constant and the vertical speed can be calculated using $v=u+at$.

Once you've found out all the values of the constants, you can differentiate $s$ and find the critical points as normal.

I should also say that treating $\theta$ as your variable is where the mistake came in. $\theta$ is fixed in this problem, you even worked out it's value! Your independent variable should be time $t$.

The energy equation is given by

$$U = mgh + \tfrac{1}{2} mv^2$$

So you obtain

$$mgh + \tfrac{1}{2} mu^2 = mg(h-8) + \tfrac{1}{2} m [2u]^2$$

whence

$$3 u^2 = 16g$$

The height is given by

$$h(t) = 8 + u_y t - \tfrac{1}{2} g t^2$$

and $h(2) = 0$, thus

$$u_y = g - 4$$

As

$$u^2 = u_x^2 + u_y^2$$

we find for point $B$

$$u_x^2 = \frac{64}{3} g - [g-4]^2$$

Note that

$$\cos(\theta) = \frac{u_x}{u}$$

so

$$\cos^2(\theta) = \frac{64 g - 3[g-4]^2}{16g}$$

The minimum velocity is $u_x$