+115 Back at it!
These questions really confuse me:
A ball is thrown from the roof of a 25 m tall building. The ball reaches a maximum hiehgt of 45 m two seconds after being thrown, and hits the ground at five seconds. Determine the equation that models the path of the ball.
Answer: y=-5x(x-4)
I just need the process
+6248 This problem is ill stated.
Unless you've left some info out there are infnite solutions for y in terms of x.
There end up being two free parameters, v0 and theta, the angle at which the throw is made with
respect to the x axis.
The ball could the thrown straight up and satisfy what's given or it could thrown at any angle and
a v0 could be found to solve the conditions listed.
Did you leave something out?
+118608 Hi Rom, I agreed with you when i first read the question but we interpreted it differently from intended.
This is a simplified school question. Perhaps we are intended to use calculus, or perhaps not.
I will do it without calculus.
The trajectory of a ball will be parabolic (pre-knowledge on my part). We are only interested in height and time.
The x does not refer to horizonal distance it refers to time and I am going to change it to a t make it less confusing.
So the ball is traveling in the path of a concave down parabola. (t,y) (time, height)
If you let the top of the building be the point (0,0) that is 0 time and 0 vertical height
then the ground will be ( 5,-25)
and the maximum height coordinates will be (2,20)
It is a concave down parabola and the t intercepts will be 0 and 4 (becasue it is symmetrical)
so the eqution must be
\(y=-a(t-0)(t-4)\\ y=-at(t-4)\\ When \;\;t=2,\;y=20\\ sub\\ 20=-a*2(2-4)\\ 20=-2a*-2\\ 20=4a\\ a=5\\ \text{So if the coordinates are set as I have described, I mean with the top of the buildng being height 0}\\ then\\~\\ y=-5t(t-4) \text{ Will model the path of the ball} \)
I hope that helps.
