Using derivatives to compute velocity if only displacement is given

Question

I am reading about the Chain Rule of derivative and encountered this problem:

If a ball is flying at an angle of π/4, what is the required velocity so it will reach 40 foot high and 350 feet horizontal distance?

The given displacement function is:

x(t) = v.cos($\theta$).t

y(t) = v.sin($\theta$).t-(16t$^2$)

Where t is time in second and v is the initial velocity.

I couldn't figure out where to start. I tried solving for t as the time when ball reaches the maximum height but I am stuck there.

Answer 1

You have that $\theta$= $\pi$/4 So you can get the cosine and sine So you just have to put that:

x(t) = v.cos($\theta$).t=350

y(t) = v.sin($\theta$).t-(16t$^2$)=40

And solve these 2 equations in 2 unknowns to get v and t

Answer 2

The vertical speed is $v\sin\theta-32 t$. You cancel it to find the time to the maximum height. Then you plug this time in the $y$ equation to find the maximum altitude.

To find the horizontal distance, solve the equation $y(t)=0$ for $t$, and plug the value of $t$ in the $x$ equation.

It is possible that you find two different values for $v$. Logically, you should pick the largest.

Answer 3

You know the max height will be when the ball starts to travel downwards again, so the rate of change of vertical velocity will be 0. Try differentiating your $x(t)$, plugging in 0 and seeing what you get for $t_1$.

For the distance, where the ball lands will be at time $t_2$ where $y(t_2)=0.$

Answer 4

You would be much more well-advised to solve this kind of problem by using the energy principle. Write the initial kinetic energy as $T=\frac{1}{2}mv^2$ and the potential energy as the top as $V=mgh$. Then, at the peak, $T=V$, and you can very easily solve $v$. Note that this concerns the vertical component of velocity because the gravity does not do anything to the horizontal component. You could write $v_{tot}=\sqrt{2}v$ to get the total initial speed becuase $v$ is the projection of $v_{tot}$ on y axis.

Then, how far the ball flies? This you can solve by concerning $t_{peak}=v/g$, i.e. gravity takes $t_{peak}$ seconds to wear out the y component of velocity. The total flying time is, of course, $2t_{peak}$ and the journey travelled $v2t_{peak}$. I would say that always use the energy principle in problems involving gravity.

You might ask can you separate kinetic energy in $x$ and $y$ components. Yes, you can. Here's the proof: \begin{equation}T_{tot}=\frac{1}{2}m|v|^2=\frac{1}{2}m\left(\sqrt{v_x^2+v_y^2}\right)^2=\frac{1}{2}m\left(v_x^2+v_y^2\right)=\frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2=T_v+T_y.\end{equation} In your ball example, I have denoted the y component as plain $T$.