# Using derivatives to compute velocity if only displacement is given

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.

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

• This is a correct method. Eliminating v, you get an equation of the form t^2 = numbers, and you use the positive root (obviously). To add some physics/intuition - note that the angle is 45 degrees - the optimum angle for maximum distance, and notice that since the horizontal distance (350) is much larger than the vertical one (40), the ball reaches this point on its way back down. – osbert Sep 4 '18 at 9:14

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: $$T_{tot}=\frac{1}{2}m|v|^2=\frac{1}{2}m\left(\sqrt{v_x^2+v_y^2}\right)^2=\frac{1}{2}m\left(v_x^2+v_y^2\right)=\frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2=T_v+T_y.$$ In your ball example, I have denoted the y component as plain $T$.

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.

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 $y(t_2)=0.$