# Homework Help: Catapult and velocity

1. Aug 31, 2009

### rebeccah78

Please,I need help with this problem....

You are launching a 200kg projectile over the castle wall and trying to hit the keep. If the catapult holds the projectile for 2 seconds before releasing what is the average force delievered by the catapult?

The keep itself is located 75 meters inside the castle wall while the catapult is 50 meters outside the castle wall. Determine the initial velocity (speed & angle with respect to the horizontal) of the projectile if it is to just clear the castle wall and impact the keep.

Remember the wall is 12 meters high.

2. Aug 31, 2009

### Staff: Mentor

Welcome to the PF. Per the Rules link at the top of the page, you must show your attempt at a solution before we can offer tutorial help. You are also supposed to use the Homework Help template that you are provided when starting a new thread here.

Show us the relevant equations, and your attempt at a solution.

3. Aug 31, 2009

### rebeccah78

*Given Information*

1.) mass of 200 kg

2.) t = 2 sec

3.) change in Y1 = 12 m

4.) X1 = 50 m, X2 = 75 m, total X of 125 m

*Formulas*

1.) (change in Y) = V(init)^2 sin^2(theta)/2g -------> theta = tan-1[(4*(change in y))/x]

2.) x = 2V(init)sin(theta)cos(theta)/g

3.) F = ma + mg

*Solution*

1.) To find angle to get projectile over wall...

theta = tan-1(change in y/x)
= tan-1(12/50)
= 13.5 degrees

2.) To find height of highest point....

tan 13.5 degrees = h/62.5 m
h = 15 m

3.) New angle...

theta = tan-1[(4*change in y)/x]
= tan-1[(4*15m)/125 m]
= 25.6 degrees

4.) Initial velocity at above angle...

V(init)^2 = xg/2sin(theta)cos(theta)
V(init) = square root of (xg/2sin(theta)cos(theta))
= square root of [(125 m(9.8 m/s^2))/2(sin25.6 degrees)(cos 25.6 degrees)
= 39.78 m/s at 25.6 degrees

5.) Average force of catapult....

a = v/t
= 39.78/2s
= 19.89 m/s

F = ma + mg
= 200 kg(19.89 m/s) + 200 kg(9.81 m/s^2)
= 5940 N