[malic]

A 3.00kg ball is dropped from a building 176.4 meters high. While the ball is falling to earth a horizontal wind exerts a constant force of 12.0 N on the ball

Questions:
a. how long does it take to hit the ground?
b. how far from the building does the ball hit the ground?
c. what is its speed when it hits the ground?

Answers:
a. 6.00 s
b. 72.0 m
c. 63.6 m/s


I don't know how to get those answer.

[tjsail]

a. You can ignore the horizontal wind. Any horizontal speed it picks up doesn't change the increasing vertical speed.
y = Vo*t + (1/2)*g*t^2
where Vo = 0 and g = 9.8 m/s^2. Solve for t.

b. Hmmm. I pictured the wind blowing parallel with the wall. If so, the answer would be right along side the wall, but to the right or left a ways compared to where it was dropped from. Make sense to you?

But for the question to make sense the windows of the building must be open so the ball is being blown away from the wall. OK, dealing with horizontal force, acceleration, and displacement:
Determine the acceleration.
F = m*a

To find distance use
x = Vo*t + (1/2)*a*t^2
where Vo = 0, and where t is from your answer a.

c. You could calculate vertical velocity and horizontal velocity separately using
Vf^2 = Vo^2 + 2*a*d
both times. As you plug in data, a is first g (=9.8 m/s^2), then the value from part b. and d is first x, then y.

But I would try conservation of energy if you have studied that. Calculate potential energy for the point it was dropped from (m*g*h) and then calculate the work the wind did (Force*distance = F*x where x is from part b.

I haven't worked these out. I'm pretty confident you will get the right answers.

[malic]

a. You can ignore the horizontal wind. Any horizontal speed it picks up doesn't change the increasing vertical speed.
y = Vo*t + (1/2)*g*t^2
where Vo = 0 and g = 9.8 m/s^2. Solve for t.

b. Hmmm. I pictured the wind blowing parallel with the wall. If so, the answer would be right along side the wall, but to the right or left a ways compared to where it was dropped from. Make sense to you?

But for the question to make sense the windows of the building must be open so the ball is being blown away from the wall. OK, dealing with horizontal force, acceleration, and displacement:
Determine the acceleration.
F = m*a

To find distance use
x = Vo*t + (1/2)*a*t^2
where Vo = 0, and where t is from your answer a.

c. You could calculate vertical velocity and horizontal velocity separately using
Vf^2 = Vo^2 + 2*a*d
both times. As you plug in data, a is first g (=9.8 m/s^2), then the value from part b. and d is first x, then y.

But I would try conservation of energy if you have studied that. Calculate potential energy for the point it was dropped from (m*g*h) and then calculate the work the wind did (Force*distance = F*x where x is from part b.

I haven't worked these out. I'm pretty confident you will get the right answers.

I still don't get c. a and be I got after you told me how but I can't find the right answer for c.

[tjsail]

I guess I didn't give you all the steps. I tried these just now and they worked.

I'll revise my advice:
c. You could calculate vertical velocity and horizontal velocity separately using
Vf^2 = Vo^2 + 2*a*d
both times. As you plug in data, a is first g (=9.8 m/s^2), then the value from part b. and d is first x, then y. The 2 times you go through the calculation for a Vf you get a vertical component and a horizontal component of velocity. These are perpendicular vectors. Combine them using the vector methods you have learned to form the resultant velocity. (You don't need to figure out the angle for this question.)

But I would try conservation of energy if you have studied that. Calculate potential energy for the point it was dropped from (m*g*h) and then calculate the work the wind did (Force*distance = F*x where x is from part b.

The energy that must be conserved is the potential energy from being up on the roof + the work done by the wind. So

potential energy + wind's work = final kinetic energy
m*g*h + F*x = (1/2)*m*v^2
Solve for v and you've got it.

[terryg752]

3.00kg ball is dropped from a building 176.4 meters high.


VERTICAL

Initial velocity = u = 0 s = -174.6
Final velocity = v = ?
acceleration = a = -g = -9.8

s = ut + 1/2 at^2

-174.6 = 1/2 (-9.8) t^2

t = 6


Final Vertical velocity =

v = u + at = 0 - 6 times 9.8 = -58.8

Magnitude of v = 58.8





HORIZONTAL

Force = ma

12 = 3 a

a = 4



time = t = 6

s = ut + 1/2 at^2

= 1/2 times 4 times (6)^2

= 72

HORIZONTAL DISTANCE = 72

u = 0, t = 6, a = 4

Final Horizontal velocity = V = u + at = 6 times 4 = 24
Final vertical velocity = calculated above = v = 58.8

Final Velocity = sqrt (V^2 + v^2) = 63.50

[terryg752]

Another interesting way of finding final velocity:

Work done by wind = 12 times horizontal distance

= 12 times 72

= 864

Potential energy of mass = mgh = 3 times 9.8 times 176.4

= 5186.16


Suppose final velocity = V

Final Kinetic Energy = 1/2 m V^2 = 864 + 5186.16

m = 3

V = 63.5