Thread: Vector kinematics projectiles thrown at an angle

The question:

The solution:

Don't you have to take into consideration the fact that the projectile was launched at an angle and was above the building for a period of time before falling to the ground? If this were the case then wouldn't the 2.42 seconds be including this period which would mean that the 12 m is in fact the sum of the building's height and the maximum height the projectile reached?

Originally Posted by iamanoobatmath

Don't you have to take into consideration the fact that the projectile was launched at an angle and was above the building for a period of time before falling to the ground? If this were the case then wouldn't the 2.42 seconds be including this period which would mean that the 12 m is in fact the sum of the building's height and the maximum height the projectile reached?

The calculation for the vertical direction is finding the vertical displacement. That is, the change in vertical position.

As it happens, the answer should be d = -11.66 m (correct to two decimal places). This makes sense because the downwards direction is negative and the upwards direction is positive. The magnitude of this displacement is the height of the building. Therefore the height of the building is 11.66 m.

whoops!

Originally Posted by tanujkush

True, you do have to do that. Since your book calculates the displacement of the projectile, the 12m computed includes the following:
1. The max height to which the projectile goes up
2. The same distance as 1 above, when the projectile comes back to the same height from which it was shot (i.e. the top of the building)
3. The height of the building.

So the 12 = 2*max height + height of building

No, this is not correct. See my previous post. Note that the top of the building is taken as the origin.