Question

URGENT!!!!!
A stone of mass m is dropped from a sheer cliff of height h. The wind exerts a force F on the stone. Assume the wind is parallel to the face of the cliff and the ground and F is constant. (Use any variable or symbol stated above along with the following as necessary: g. For all vectors, enter the magnitude.)

1)If t = 0 corresponds to the moment the stone is released, at what time t does it strike the ground?

2)Find an expression in terms of m and F for the acceleration ax of the stone in the horizontal direction (taken as the positive x-direction).

3)How far is the stone displaced horizontally before hitting the ground?

4)Find the magnitude of the stone's acceleration while it is falling.

5)What If? Recalculate your answers to parts (a) through (d) for the case where the wind instead blows at an angle θ below the horizontal. (Use the following as necessary: m, g, F, h and θ.)

Answer 1

Answer:

1.

$t = \sqrt(2h/g)$

2.

$ax = F/m$

3.

$x=\frac{Fh}{mg}$

4.

$a=\sqrt(F^2/m^ + g^2)$

5.

(a)

$t = \sqrt\frac{2h}{g+{Fsin\theta}{m}}$

(b)

$ax= \frac{Fcos\theta}{m}$

(c)

$x=\frac{hFcos\theta}{m(g+\frac{Fsin\theta}{m}}$

(d)

$a=\sqrt((\frac{F^2cos^2\theta}{m^2}+(g+\frac{Fsin\theta}{m})^2)$

Explanation:

1.

$h = 0.5gt^2$

Therefore, $t = \sqrt(2h/g)$

2.

$ax = \frac{F}{m}$

(3)

$x = 0.5axt^2 = \frac{0.5F/m}{2h/g}  = \frac{Fh}{mg}$

(4)

$a = \sqrt(ax^2 + g^2) = \sqrt(F^2/m^ + g^2)$

5

(a)

$h=0.5(g+\frac{Fsin\theta}{m}$

$t = \sqrt\frac{2h}{g+{Fsin\theta}{m}}$

(b)

$ax= \frac{Fcos\theta}{m}$

(c)

$x=0.5axt^2$

$x= 0.5\frac{Fcos\theta}{m}t^2$

$x=\frac{hFcos\theta}{m(g+\frac{Fsin\theta}{m}}$

(d)

$a=sqrt(ax^2+ay^2)=\sqrt((\frac{F^2cos^2\theta}{m^2}+(g+\frac{Fsin\theta}{m})^2)$

Answer 2

Acceleration is the rate of change of velocity with respect to time. Stone's acceleration is $a = \sqrt{(\dfrac{F}{m})^2 + g^2}$.

What is acceleration?

Acceleration is the rate of change of velocity with respect to time.

A.) We know that according to the second equation of motion for the vertical moment we can write the equation as,

$S = ut + \dfrac{1}{2}gt^2$

Now, since the initial velocity of the stone is 0, and the S the distance that will be covered by the stone is the height of the cliff h. therefore,

$h = \dfrac{1}{2}gt^2\\\\t= \sqrt{\dfrac{2h}{g}}$

B.) we know that the force is the product of mass and acceleration, therefore, the acceleration of the stone can be written as,

$F = m \times a_x\\\\a_x = \dfrac{F}{m}$

C.) We know that according to the second equation of motion for the horizontal moment we can write the equation as,

$S = ut + \dfrac{1}{2}at^2$

Now, since the initial velocity of the stone is 0,

$x = \dfrac{1}{2}a_xt^2$

We know the values of $a_x$, also, the value of t, therefore,

$x = \dfrac{1}{2} \times \dfrac{F}{m} \times (\sqrt{\dfrac{2h}{g}})^2\\\\\\x = \dfrac{1}{2} \times \dfrac{F}{m} \times {\dfrac{2h}{g}}\\\\\\x =\dfrac{Fh}{mg}$

D.) The magnitude of the stone's acceleration while it is falling.

Stone's acceleration while it is falling, is the resultant of the vertical and the horizontal acceleration that is working on the stone,

$a =\sqrt{a_x^2 + a_y^2}\\\\a = \sqrt{(\dfrac{F}{m})^2 + g^2}$

E.) If the force is applied from an angle θ, then,

1.

$h = \dfrac{1}{2}(g+ \dfrac{F\ \Sin \theta}{m})\\\\t= \sqrt{\dfrac{2h}{(g+ \dfrac{F\ \Sin \theta}{m}}}$

2.

$a_x = \dfrac{F\ Cos \theta}{m}$

3.

$x = 0.5a_xt\\\\x = 0.5 \dfrac{F\ Cos \theta}{m}t^2\\\\x = \dfrac{h\ F\ Cos\theta}{m(g+\dfrac{F\ Sin\theta}{m})}$

4.

$a=\sqrt{a_x^2+a_y^2}\\\\a = \sqrt{(\dfrac{F\ Cos \theta}{m})^2+(g+ \dfrac{F\ \Sin \theta}{m})^2}$

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