Posted by Max on Thursday, February 11, 2010 at 4:06pm.
You are answer grazing, I think.
What is your thinking on this?
I am not answer grazing. I just don't know how to do these couple problems at all. I have no idea where to start.
Let Vi be the velocity. The vertical component then is Vi sin24. The horizontal component is Vi Cos24
In the horizontal, you know distance, time, solve for Vi.
Then, knowing Vi, you know the vi vertical, time in air. Solve for hi
hf=hi+vivertical*t-4.9 t^2
Finally, max height occurs when vivertical is zero. Find that time, then find h.
What is the equation you use, knowing distance and time, to get Vi?
Ok I figured out Vi. Now I don't get what you put in for Hf because you can't have two variables to find out Hi?
I'm still not able to find Hi with the given equation. Could you possibly explain more on how to get the initial height?
From the edge of the rooftop of a building, a boy throws a stone at an angle 24.0° above the horizontal. The stone hits the ground 4.40 s later, 104 m away from the base of the building. (Ignore air resistance.) Find the initial velocity of the stone and magnitude and direction above the horizontal. Find the initial height h from which the stone was thrown. Find the maximum height H reached by the stone.
a = -9.8
U = Speed cos 24 = S cos 24
Vi = Speed sin 24 = S sin 24
V = Vi - 9.8 t
h = Hi + Vi t - 4.9 t^2
First find the constant horizontal speed U = 104/4.4
then the initial speed S = (104/4.4)/cos 24
then the initial velocity up Vi = S sin 24
h = 0 at ground so
0 = H +Vi t -4.9 t^2
but t = 4.4 so solve for H
also
at the top, v = 0
so
0 = Vi - 9.8 t at top
that gives you t at top
use that t in
h = H + Vi t -4.9 t^2