How long does the ball take to hit the ground?

[princesspriya]

Homework Statement

a 3 kg ball is dropped from rest from the roof on a building 176.4 m high. While the ball is falling, a horizontal wind exerts a constant force of 12 N on the ball?
Part 1) How long does the ball take to hit the ground?
Part 2) How far from the building does the ball hit the ground?
Part 3) When the ball hits the ground, what is its speed?

Homework Equations

The Attempt at a Solution

y=176.4
Vf=?
Vi=0
a=9.81

Part. 3) Vf^2=Vi^2+2ay
Vf= 58.83 m/s

Part. 1) Vf=Vi+at
t=5.99

I don't understand how to find Part 2 because its falling under free fall acceleration so wouldn't it just fall straight down? and the equation to find the x=ViT but the Vi=0 so would'nt the ball just fall straight down instead of falling far away from the building?

 

[Snazzy]

Draw a free-body diagram and see what forces are acting on the ball. They give you the wind resistance for a reason.

 

[Moetasim]

I would like to clarify that there are a few assumptions that need to be made in order to accurately answer the questions:

1. The ball is a perfect sphere with uniform density.
2. The wind force is the only external force acting on the ball.
3. The wind force does not change during the ball's fall.
4. The wind force acts horizontally and does not affect the vertical motion of the ball.

With these assumptions in mind, here is the solution:

Part 1) How long does the ball take to hit the ground?

We can use the equation y = y0 + v0t + 1/2at^2 to solve for t, where y0 is the initial height, v0 is the initial velocity, a is the acceleration due to gravity, and t is the time.

Substituting the given values, we get:

y = 176.4 m
v0 = 0 m/s
a = 9.81 m/s^2

Solving for t, we get t = 5.99 seconds.

Therefore, the ball takes 5.99 seconds to hit the ground.

Part 2) How far from the building does the ball hit the ground?

Since the wind force does not affect the vertical motion of the ball, the ball will fall straight down. However, the wind force does affect the horizontal motion of the ball.

We can use the equation x = x0 + v0t + 1/2at^2 to solve for x, where x0 is the initial position, v0 is the initial velocity, a is the acceleration due to gravity, and t is the time.

Substituting the given values, we get:

x = x0 + v0t + 1/2at^2
x0 = 0 m
v0 = 0 m/s
a = 0 m/s^2

Solving for x, we get x = 0 m.

Therefore, the ball hits the ground directly below the building, at a distance of 0 m.

Part 3) When the ball hits the ground, what is its speed?

We can use the equation vf^2 = v0^2 + 2a(y-y0) to solve for vf, where v0 is the initial velocity, a is the acceleration due to gravity, y is the final height (which is 0 m), and y0 is