Help with Constant acceleration question?

Question

Can anyone help - I am ok when calculating something dropped, but I don't know what I need to find out in the following;

A particle is thrown into the air at a speed of 15m/s from a height of Hm and reaches the ground in 4secs. Find the greatest height it reaches and the value of H.

Answer 1

Start by finding the time it reaches zero speed.

v = v0 + at ---> 0 = 15 -9.8 t ---> t =15/9.8 ~ 1.5 sec

So it takes t = 4 - 1.5 = 2.5 sec to hit the ground from it's high point. The high point is:

y = 1/2at^2 = 1/2*(-9.8)*(2.5)^2 = -31.25 m from high point. Minus sign is because measuring down which I've chosen tobe negative direction. So max height is 31.25 m above ground.

Now when it passes H it is moving at 15 m/s. Further it has been falling for 1.5 sec. So

y - H = 1/2at^2 = -1/2*(9.8)*(1.5)^2 = 11.25 m

Then H = y - 11.25 = 31.25 - 11.25 = 20 m

Answer 2

First of all calculate the maximum height. The stone will decelerate until its speed is zero - ie when it reaches maximum height.
Using the formula v^2 = u^2 + 2*a*s we have

0 = 15 -2g*s
where g is the acceleration due to gravity (look it up) and s is the height it reaches and so s = 15/2*g

So the maximum height is H + 15/2*g

The time taken to reach this height is given by s = 1/2 gt^2
You have found s (15/2*g) so you can find t.

So you now know how long it takes to get to the top. Subtract this from 4 to get how long it takes to reach the ground from the top.
Use this value of time again in the formula s = 1/2*gt^2 to get the value of s which will be the total height from the ground (which is one of the values you want).
Subtract 15/2*g from this to get H (which is the other value you want).

Now all you have to do is work out the sums!