Question

An object of mass m is dropped at t=0 from the roof of a building of height h. While the object is falling, a wind blowing parallel to the face of the building exerts a constant horizontal force F on the object. a) At what time t does the object strike the ground? b) Find an expression in terms of m and F for the acceleration in the horizontal direction. c) How far is the object displaced horizontally before hitting the ground? Answer in terms of m, g, F, and h. d) Find the magnitude of the object’s acceleration while it is falling, using the variables F, m, and g.

Answer 1

I'll assume we can ignore wind-resistance. There are two constant forces that act on the object, gravity in the y and wind in the x, therefore, we can use our kinematic relationships for constant acceleration.

For part a, we are only interested in the vertical motion of the particle. Noting that the distance is travels in the horizontal direction has no impact on the time it takes to hit the ground. Let's solve the following kinematic equation for t\[h = v_o t - {1 \over 2} g t^2\]Since the object is dropped, \(v_o = 0\).
\[t = \sqrt{~2 h \over g}\]

For part b, let's recall that the force is constant, therefore from Newton's Second Law\[F_w = m \cdot a_x\]

For part c, we will follow a similiar methodology as in part a. Again, recalling that the force is constant and as such the acceleration will be constant. Also, note that the horizontal motion is not affected by the vertical motion. The horizontal motion however is limited by the time it takes the object to fall the height of the building. We can again use the following kinematic equation\[d_x = v_{o,x} t + {1 \over 2} a_x t^2\]Note that \(v_{o,x} = 0\), \(a_x = {F_w \over m}\), and \(t = \sqrt{2 h \over g}\)\[\therefore d_x = {1 \over 2} \left( F_w \over m \right) \left(2h \over g \right)\]


Part d is straight forward, the magnitude of the acceleration can be expressed as\[a = \sqrt{a_y^2 + a_x^2}\]Note that \(a_y = g\) and \(a_x = {F_w \over m}\)\[\therefore a = \sqrt{g^2 + \left (F_w \over m \right)^2 }\]

Answer 2

@eashmore
For part a, you used the kinematic equation \[h=v _{ox}t-(1/2)g t^{2}\].
However, you also said that \[v _{ox}=0\] so that would mean h is negative. Since you cant take the square root of a negative value, how did you solve for t?

Answer 3

I forgot to change the sign. I considered the downward direction positive. Sorry about the confusion.

Answer 4

@eashmore Could you write out what you mean please? I'm not sure which sign you're referring to.

Answer 5

I'm considering the positive y-direction to be downwards, this is opposite the usual convention where upwards is positive.

Therefore, the objects falls a positive displacement and gravity has a positive sign, thus\[h = {1 \over 2} g t^2\]

If we take the usual sign convention (upwards being positive) note that the object has a negative displacement, thus\[- h = -{ 1 \over 2} g t^2\]