Sorry Charlie, ain't gonna happen.

Quoted: You will hit the object and it will bounce back and break your nose. Dont do it.

Saw that in HS... Anvil, guy pulling out the object to be hit, and a BFH.

=

Broken nose

Quoted: You will hit the object and it will bounce back and break your nose. Dont do it.

Saw that in HS... Anvil, guy pulling out the object to be hit, and a BFH.

=

Broken nose

Quoted: Quoted: <original comment edited out> Edited my post out. Jumped the gun and acted like a douchebag. My apologies Sparsky. I can understand the comment. We already have all the answers, we're just tring to figure out how to get those answers. I have only taken high school physics. My wife is the one in college, I'm tring to help her out with aid from my brothers on arfcom. the questions i didn't post were. What's the time at the point at which the ball is directly at ann's brother. we have formula w/ time of .50sec Where is the ball at Ann's bro (above/below/hit) Above w/ .13m clearance how long does it take for the ball to reach it's peak and total time for trip? time .59sec   total time 1.18sec Total length traveled by ball 4.83m Now I'm just tring to help her/me understand how they got all of this with all these lettered formulas. That and how would we get these answers with out all the answers given. Edited to add: No worries mate.... it happens to the best of us.

Ok kinematics can be broken down into three simple and easy to remember equations.


Velocity Final =  Velocity Initial + acceleration*time
Vf=Vi+a*t

(Velocity Final)^2= (Velocity Initial)^2 + 2*(Acceleration)*(displacement)
Vf^2=Vi^2 +2*a*s

Final Displacement= Initial Displacement + Initial Velocity* Time  + 1/2*acceleration*(time)^2
Sf= Si + Vi*t + 1/2*a*t^2


So. In this case, what are we trying to accomplish? We want to know if the ball is ever at x=4, y=1..
So we use our third equation, but we first need to break things down. Now we have a vector, it says that the ball travels at a velocity of 10 m/s, at an angle of 35 degrees from the horizontal. We want to break this down into x and y components. We must fall back on our knowledge of trigonometry.

The cosine function is cos(angle)= (adjacent leg of angle) / (hypotenuse of angle). So call the angle the greek letter theta, cos(theta)=adjacent/hyponetenuse. We know the hypotenuse, which is 10 m/s, we use algebra to solve for the adjacent leg. So the horizontal velocity is the adjacent leg and it is equal to 10 m/s * cos(35), which is about 8.2m/s.

Sine(theta)=opposite/adjacent
so y velocity is 5.73.

So intial velocity in the x direction is about 8.2 m/s. Now that we know this we can plug it into our third equation. We want to know which t satistifies the 4 meter horizontal condition. Remember there is no acceleration so that gets rid of the squared term, and no initial displacement

So, 4meters= 0 meters + 8.2*t +.5*0*t^2

4m=8.19m/s*t

meters cancel out, leaving only seconds
t=.49 seconds

This means that the ball can only be 4 meters away at roughly half a second. We need to know where it is hieght wise at that time. Use the third equation again
y= 0 + 5.73*(.49)+.5*-9.8*(.49)^2= 1.63

This says that at .49 seconds the ball is 1.63 meters high, so it misses the target by .13 meters.

The ball reaches its peak when vertical velocity is zero,

0=5.73-9.8*t
-5.73=-9.8
t=.59

so at t=.59 seconds the ball is at its peak height.

and to find the total distance traveled you simply multiply total flight time by velocity.(this is a seperate computation, it is easy but im too tired and im going to bed)