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integra2k20
Oct4-06, 05:18 PM
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?

civil_dude
Oct4-06, 05:31 PM
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.

integra2k20
Oct4-06, 05:46 PM
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...

integra2k20
Oct5-06, 03:30 PM
i asked my teacher about this today, he couldnt figure it out either...

DV10
Mar15-10, 04:09 PM
2.3 degrees or 78 degrees?
like civil said,it can be calculated..just check again.