I am having trouble wrapping my head around this problem. The big picture is that i have to calculate the initial velocity v= needed for a soccer ball to cross a goal line.
this is a homework problem, im not looking for someone to do the problem, just a push in a good direction...
known information:
(x0,y0) = where the ball is kicked from = (0, 2)
the angle at which the ball is kicked is 15 degrees
the goal is 60ft away
the location the ball must cross the plane is (60, 7)
v(t) = <Cos[x], Sin[x]>
considering only gravity acting on the ball.
r(t) = <u0t + x0, -16t^2 + v0 + y0>
if i set the u0t = 60+(0) and 7=-16t^2+v0+2; then solve for t I get:
t=60/u0 and t=sqrt(v0-5/16)
but then im not sure exactly what to do? should i plug t back into the position function? if i do that then i have two equations with 3 unknowns...
-16(60/u0)^2+sCos[15](60/u0)+2
-16((v0-5)/16)^2+sSin[15]((v0-5)/16)+2
where s is the speed of the ball.
not sure what to do here...
Your formula for $r(t)$, the position of the ball, is wrong, since you left out a $t$ with $v_0$. The correct formula is
$$r(t)=<u_0t+x_0,\ -16t^2+v_0t+y_0>$$
Let $v$ be the magnitude of the ball's initial velocity. Then the horizontal and vertical components of the initial velocity are
$$u_0=v\cos 15°, \quad v_0=v\sin 15°$$
Substituting the information into the horizontal and vertical components of $r(t)$ we get
$$60=(v\cos 15°)t+0, \quad 7=-16t^2+(v\sin 15°)t+2$$
We now have two equations in the two unknowns $t$ and $v$, which you should be able to solve. One of the equations is quadratic, but it's not all that hard. This is the kind of problem I assign each year to my 12-th grade physics class, which makes my students complain that it is more of a math class than a physics class.