Minimum speed using calculus

Question

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.

Answer 1

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 $t$.

Answer 2

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 $u\cos\theta$.

Answer 3

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

$$ 3 u^2 = 16g $$

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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 $B$

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

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Note that

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

so

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

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The minimum velocity is $u_x$