Posted by **Jeremy** on Sunday, January 24, 2010 at 2:34pm.

This is a pretty simple physics problem, but I do not know how to go about it.

How do I:

Show that the time required for a projectile to reach its highest point is equal to the time for it to return to its original height?

- Physics -
**drwls**, Sunday, January 24, 2010 at 3:10pm
The maximum height is where V = 0.

The acceleration rate is -g, in either direction.

The initial speed is the same as the final speed (because of conservation of energy.)

The time it takes to decelerate from initial speed V to 0 is V/g.

The time it takes to accelerate from 0 to -V is also V/g

This works whether the projectile goes straight up or not. V represents the vertical velocity component.

## Answer this Question

## Related Questions

- Physics - A ball is thrown vertically upward, starting at y = 0, with a velocity...
- Physics - "A projectile is launched at 517m/s at 50 degrees from the horizontal ...
- physics - a projectile is thrown from the ground with an initial velocity of 20....
- Physics - A projectile is launched at an angle of 34.0o above the horizontal. ...
- physics - A projectile is launched from ground level at 37.0 m/s at an angle of ...
- physics - A projectile is launched from ground level at 37.0 m/s at an angle of ...
- Physics - A projectile is launched with an initial speed of 40m/s at an angle of...
- Physics - A projectile is launched with an initial speed of 80.0m/s at an angle ...
- physics - a projectile is thrown from the ground with an initial velocity of 20....
- physics - A projectile is launched with 30m/s at an angle of 60degree with d ...

More Related Questions