# Finding the angle of a dive bomber droping a bomb.

## Homework Statement

A dive bomber has a velocity of 275 m/s at an angle θ 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 3.35 km. Find the angle θ?

## The Attempt at a Solution

I have tried this like 6 different ways and they are all wrong. I'm confused if they want the instal angle or the angle form the bomb so Please help me.

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CWatters
Homework Helper
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Is there a diagram or more info to explain what this bit means..

"The magnitude of the displacement from the point of release of the bomb to the target is 3.35 km"
Is that the straight line distance from the point where the bomb is released to the target OR just the horizontal displacement?

You need to calculate the horizontal displacement if thst's not what they have given you. Shouldn't be a problem you know two sides of a right angle triangle. Calculate the third if necessary.

The bomb will initially follow the plane down at the angle θ but then curve away downwards due to gravity.

I would write two equations:

Vertically - an equation for the time it takes for the bomb to fall from 2.15km to the ground. It will start falling with an initial vertical velocity that depends on θ and the velocity of the plane and then acceleration due to gravity. One of the standard equations of motion should do as a starting point.

Horizontally - an equation for the time it takes the bomb to travel the horizontal distance from release point to target. This is simpler. The horizontal velocity can be assumed constant and depends on θ and the velocity of the plane. You were either told or calculated the horizontal displacement. Time = distance/velocity.

The time taken to move vertically is equal to the time taken to move horizontally so you can equate these two equations. Then I think it should be possible to solve for θ