Interesting Projectile Motion problem

[Mattowander]

Homework Statement

A person throws a ball off a roof that is 8 meters high with an angle of 45 degrees from the positive x axis. The ball is supposed to land in the center of a square that is centered 9 meters away from the base of the roof. With which initial velocity must the person throw the ball so that it lands from it is supposed to?

Homework Equations

Vi = unknown
Vx = 9/Ttot
Vx = Vi cos (45)
Vy = Vi sin (45)
h = 9 meters
y = distance from the initial height of the ball to the height that the ball peaks.
Ttot = the total time the ball is in there air.

The Attempt at a Solution

Ttot = t1 + t2 where t1 = the time it takes the ball to peak and t2 = the time it takes the ball to fall from that maximum height.

t1 = -Vy/g
t2 = sqrt ( (2h + 2y)/g)

t3 = -Vy/g + sqrt ( (2h + 2y)/g) = 9/Vx

When replaced with everything I'm given, I get the following equation. Keep in mind we're trying to solve for Vi.

t3 = -Vi sin(45)/g + sqrt ( (2(8) + 2(-Vi sin (45)/2g)/g) = 9/(Vi cos (45) )

From here, I'm not sure how I can solve for Vi. Is this problem impossible to solve with what I'm given or am I just making it a lot harder than it needs to be? I'd appreciate any help ya'll can give me!

 

[Lancelot59]

I'd just use the component method.

This is what you end up with

V_x = V_ocos(\Theta)
V_y = V_osin(\Theta)

Because the angle is 45^o, what can you assume about the X and Y components of the velocity?

From that you can use the kinematics formulas to mess around with the stuff you have and find other information.

For instance:

d=V_ot+(1/2)at²

You know what the final vertical displacement will be, and from this you can substitute and solve.

Or you could try using other kinematics equations.

Just remember that X only has one equation: D_x=V_xt

 

[Mattowander]

No you don't know what the final vertical displacement will be because you don't know the initial velocity.

Yes I know the velocity components of Vi are the same.

Did you look at the problem?

 

[Lancelot59]

You said it needs to land in a square at ground level from a 8 METER high roof. So your displacement is -8.

 

[Mattowander]

No because if it was initial velocity UPWARD, the initial height will be greater than 8 meters. The additional height will be equal to Vy^2/2g.

Since I don't know Vy without knowing Vi, I don't have the complete vertical displacement.

 

[lewando]

I agree with Mattowander's initial assertion that he is making this harder than it needs to be. Setting up the x-y origin at the edge of the roof (implied by Lancelot), concentrate on solving the following equations at the specific time T when the projectile is at (X, Y) = (9, -8)

X = vxi * T [x component of motion]

Y = (vyi * T) -9.8 T^2 [y component of motion]

vxi = vyi

 

[lewando]

Sorry, forgot the 1/2
Y = (vyi * T) -(9.8 T^2)/2 [y component of motion]

 

[essecks]

No because if it was initial velocity UPWARD, the initial height will be greater than 8 meters. The additional height will be equal to Vy^2/2g.

Since I don't know Vy without knowing Vi, I don't have the complete vertical displacement.

http://en.wikipedia.org/wiki/Displacement_%28vector%29" is the distance from A to B, not the distance you traveled in getting there.