How to Calculate Initial Speed and Impact Speed of a Vertically Thrown Stone?

[markosheehan]

a stone us thrown vertically upwards under gravity with a speed of u m/s from a point 30 metres above the horizontal ground. the stone hist the ground 5 seconds later.

(i) find the value if u
(i)find the speed with which the particle hits the ground.

can someone help me work this out. i tried to divide it into 2 different parts. from when the stone is thrown up to when it gets to v=0 and then from when it drops back down to earth. i tried to use uvast equations and i could say the 1st bit has time t and the next bit has 5-t time but the intial speeds are not the same so i don't know what to do

 

[Evgeny.Makarov]

You can use the fact that a stone dropped from rest travels distance $\dfrac{gt^2}{2}$ and has speed $gt$ after falling for $t$ seconds (or, conversely, a stone thrown up vertically with speed $gt$ reaches the maximum height after time $t$ and travels distance $\dfrac{gt^2}{2}$). Indeed, the speed increases by $g$ each second, and $\dfrac{gt^2}{2}$ is the area under the graph of speed plotted over time, which is distance.

Suppose, as you do, that the stone took $t$ seconds to reach its maximum height. Then it flew $\dfrac{gt^2}{2}$ meters over the initial position. The distance it traveled on the way down is $\dfrac{g(5-t)^2}{2}$. This allows forming an equation in $t$. Once you know $t$, you can find $u=gt$.

Another way to solve this is to use the fact that a stone thrown vertically upward with speed $u$ from height $h_0$ is at height $h(t)=h_0+ut-\dfrac{gt^2}{2}$ after time $t$. You can equate $h(t)$ to 0 and find $u$ from there.