At what velocity v do you have to throw the stone

Question

Suppose you are standing on the ground and throwing a stone straight up in the air. Let assume, for simplicity, that the gravitational constant is 10 meters per second per second. Neglecting air resistance, the stone's distance from the ground, in meters, after t seconds will be h(t)=−5t^2+vt.

At what velocity v do you have to throw the stone so that it reaches a height of exactly 9.8 meters before it starts to fall back towards the earth?

I have tried to derivate the function, but still stands still on this.

Answer 1

Your $v$ is the initial velocity, which would usually be written $v_0$ to distinguish it from the instantaneous velocity $v$. Using that, you also have $v=v_0-10t$ The highest point will come when $v=0$, so you can get $t$ and substitute in.

If you want to take the derivative, you have $h'(t)=v(t)$ so the result is the equation above.

Answer 2

It's position as a function of time, $p(t)= ut - \frac12 gt^2$.

On differentiation, we obtain it's velocity, $v(t)=u-gt$. At it's highest point $v=0$, so $t=\dfrac{u}g$. Substituting this for $t$ in $ut - \frac12 gt^2=9.8 \text{ metres}$, we obtain a linear equation in $u^2$.

Answer 3

You are close, you must evaluate the function at the maximum height, or when the derivative equals zero. There is a very nice form for quadratic function when that time occurs, it is: $$ t^*=-\frac{b}{2a}. $$ So $$ t=\frac{v}{10}\Rightarrow 9.8=-5\frac{v^2}{100}+\frac{v^2}{10} $$ and solve.