Question

Pls help if u can!!! i hate physics ughhh

a model rocket is launched horizontally from the same spot on the top of the 30.0 meter-tall building, but this time with an initial velocity of 22.0 meters per second. Predict if the rocket will hit the other 30.0-meter-tall building that is 40.0 meters away on its upper half or lower half, and perform a calculation to support your claim.


would I use the horizontal velocity equation????? ugh

Answer 1

The rocket hit the top half of the building.

Step-by-step explanation:

Here is the formula for displacement

`x(t)=(at^(2) )/(2)vt`

Because the projectile is being launched at zero degrees, all of its velocity is in the horizontal direction and so there is no initial velocity for the fall, therefore

`h=(at^(2) )/(2)`

`2h=at^(2)`

`(2h)/(a)=t^(2)`

`\sqrt(2h)/(a) =t`

Now that we have the time taken for the object to fall, we can use it to find d by applying the displacement formula again. In this case we will remove the initial term because the acceleration is zero in the horizontal direction. Therefore

`d=vt`

`d=v*\sqrt`

`d=22*\sqrt`

`d=54.41`

If there wasn't a building in the way the displacement in the x direction is 54.41.

`y=(tan0*40)-[(-9.81)/(2(22*cos0)^(2))]40^(2)`

`y=16.21`

Like you said the building is 30 meters tall. You were looking to find out is it was going to hit the top or bottom half.

`15 < 16.21`

So it hit the top half of the other building.