Question

An object is thrown from ground level at a velocity of 25.0 m/s at an angle of 20.0° above the horizontal. If there is a building 25.0 m away, at what height above the ground will the object hit the building?

Answer 1

@Hero

Answer 2

@wio

Answer 3

Since the velocity is at angle 20 to the horizon, you need to find the components of the velocity.

So in the horizontal direction, you have: \[25\sin 20= v _{x}\]. In the vertical direction you have:\[25\cos 20=v _{y}\]

Now you need to find the time it takes to get from the launch point to the building.

So the distance to the building is equal to the velocity in the horizontal direction times the time to get to the building.

\[25=(25\cos 20)(t)\]

So t=1.06sec

Answer 4

To find the height at which the ball hits the building, we need to use the vertical velocity times time and subtract the 1/2 acceleration of gravity times time squared.

\[height=(25\cos20)(1.06)-1/2(9.8)(1.06^{2})\]

height=24.9 -5.5=19.4 meters

Answer 5

Thanks but I am having trouble visualizing what the Vx and Vy represent

Answer 6

Think of a right triangle with the legs along the x and y axis. So Vy is the y axis and Vx is the x axis.

Answer 7

But doesn't Vy have initial and final whereas Vx is always constant?

Answer 8

So shouldn't there be 2 values for Vy?

Answer 9

Vx is going to hit the building in 1.06 seconds. And no there are not two Vy values

Answer 10

Ok I have done the question myself and my answer was 3.55m instead because I used -9.8 since I always knew gravity of fall object is negative.

Answer 11

That was for the last equation

Answer 12

Well I used a negative acceleration. Do you see the minus sign in front of the 1/2?

Answer 13

Do you understand where the 1/2 is coming from?

Answer 14

kinematics equation?