## How do i start here?

A dive bomber has a velocity of 275 m/s at an angle (theta) below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.45 km. Find the angle (theta).

OK, i'm not seeing a way to solve this without being given the time (t) it took for it to hit the ground, am i missing something here?
 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
 You know the time. If it is released from 2.15 km, and it is subject to g, simply figure out how long it takes to reach the ground. Ignore the horizontal portion for now.

 Quote by civil_dude You know the time. If it is released from 2.15 km, and it is subject to g, simply figure out how long it takes to reach the ground. Ignore the horizontal portion for now.
but the plane's velocity isnt horizontal....therefore, the time to reach the ground cant be solved...