# Thread: Initial velocity

1. ## Initial velocity

A football is kicked from the ground with an initial velocity of 75 ft/sec at an angle of 28 degrees from the ground. On the ball's descent, it strikes the ten foot high crossbar of the goal post. How far is the goal post from the point where the kick was made? The acceleration due to gravity is 32ft/sec^2.

I haven't the foggiest idea of how to figure this out. I've looked over and over again through the text...

2. Originally Posted by kimberlyd1020
A football is kicked from the ground with an initial velocity of 75 ft/sec at an angle of 28 degrees from the ground. On the ball's descent, it strikes the ten foot high crossbar of the goal post. How far is the goal post from the point where the kick was made? The acceleration due to gravity is 32ft/sec^2.

I haven't the foggiest idea of how to figure this out. I've looked over and over again through the text...

Since $\displaystyle v_0=75$ we can decompose the vecloity into its components

$\displaystyle v_{x0}=75\cos(28^\circ),v_{y0}=75\sin(28^\circ)$

$\displaystyle v_y(t)=\int (-32) dy= -32t+c \implies v(0)=75\sin(28^\circ)=$
$\displaystyle -32(0)+c \iff c=75\sin(28^\circ)$

$\displaystyle v_y(t)=-32t+75\sin(28^\circ)$

Now position is the integral of velcoity and $\displaystyle y(0)=0$

$\displaystyle y(t)=\int v_y(t)dt=-16t^2+(75\sin(28^\circ))t+c$

$\displaystyle y(0)=-16(0)^2+(75\sin(28^\circ))(0)+c 0=c$

$\displaystyle y(t)=\int v_y(t)dt=-16t^2+(75\sin(28^\circ))t$

setting the above function $\displaystyle y(t)= 10$ (the height of the goal post) will give you the time you need. The velocity is constant in the x direction (no acceleration) so the distance is $\displaystyle v_{0x}\cdot t$