For archery practice, a knight's squire drops sandbags from a 12.0 metre tower. At exactly the same time the sandbag is dropped, the knight shoots an arrow up at the sandbag from the base of the tower. If the arrow strikes the sandbag at 1.1 seconds, calculate;
a) how far from the ground the arrow strikes the sandbag
b) the arrow's initial velocity
I chose vf^2 = vi^2 + 2g*t equation, ended up with this...
vf = vi + g*t
d = 1/2(vi + vf)*t
d = vi*t + 1/2g * t^2
The Attempt at a Solution
= 0 + 2(-9.8)(12)
= (the square root of) 235.2
Obviously incorrect. The biggest problem I'm having is finding the right equation to use. Where exactly do I go from here?