How much is does it take for the stone to reach the bottom?

Question

So a stone is thrown vertically upward with a speed of 10.0 m/s from the edge of a cliff 65.0 m high.

So, just making sure I have all my variables right

Vinitial: 10 m/s

V: 0 m/s

g: -9.8 m/s^2

y= 0 m

y initial: 70.10 m

I got 70.10 m by doing 10^2/2(9.8)= 5.10 and added that to the cliff's height because of the ball being thrown up in the air so that equals 70.10 m... is that correct?

Also, I had to find how much later does it reach the bottom and tried using the quadratic equation for 1/2gt^2 + Vinitial(t) + y initial

I got 2. something seconds but that wasn't the right answer.

It was 4.8 s. What am I doing wrong???

Answer 1

Hint:

$$y = y_0 + v_y t + \dfrac{1}{2}a_y t^2 =65+10t-\dfrac{1}{2}gt^2=0.$$

Use the quadratic formula.

Answer 2

Hint: Use the equation of motion $x(t)=x_0+v_0t+\frac 12at^2$.

We know that $a$ is $g$, $v_0$ is $10$m/s and $x_0=65$m, because the stone is thrown from the edge of the cliff. Note that we are interested in $x(t)$ to be the position at the bottom, this is $0$. Use the equation to solve for $t$.