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Question

A stone with a weight of 4.25 N is launched vertically from ground level with an

initial speed of 18 m/s and the air drag on it is 0.25 N throughout the flight.



(a) Find the maximum height reached by the stone.

(b) Determine the speed of the stone when it hits the ground.

Answer 1

As a convention, I will consider upward direction as positive and downward direction as negative.

Let

g = -9.81 m/s² be gravitational acceleration (downward, so negative)

W = -4.25 N be the weight of the stone (downward, so negative),

m = W/g be the mass of the stone,

x(t) be the position of the stone at a time t

x(0) = 0 be the initial position of the stone,

v(t) be the velocity of the stone at a time t,

v(0)= 18 m/s be the initial velocity (upward, so positive),

t₁ be the time when stone reaches maximum height, that is when v(t₁ ) = 0,

D = 0.25 N be the absolute value of the drag (downward, when the stone is

going up, and upward when the stone is going down),

F be a symbol denoting force in general.

I will assume that you are familiar with the following two equations,

which can be derived from Newton's Second Law:

v(t) = (F/m)t + v(0) (1)

x(t) = (F/2m)t² + v(0)t + x(0) (2)

(a)

To determine x(t₁ ) plug into equations (1) and (2) the information we have about t₁ .

During the upward flight, both the weight and drag act downward, so

F = W - D = -4.5 N

0 = (F/m)t₁ + v(0) (3)

x(t₁ ) = (F/2m)t₁ ² + v(0)t₁ + 0 (4)

From (3) express t₁ = -(m/F)v(0)

and plug it in into (4):

x(t₁ ) = (F/2m)(m/F)²v(0)² - (m/F)v(0)²

= (m/2F)v(0)² - (m/F)v(0)²

= -(m/2F)v(0)²

= -(W/2gF)v(0)²

Substituting actual numbers:

x(t₁ ) = -(-4.25/(2×(-9.81)×(-4.5)))×18 = 15.6 m

(b)

Let v₂ be the speed (to be calculated) when the stone hits the ground.

We could use equations (1) and (2), but it is actually

easier to use conservation of energy:

The potential energy at the top gets converted into the kinetic energy at the bottom plus

the work of drag:

P = K + L, (5)

where

P = mgx(t₁ ) is the potential energy of the stone at the top,

K = mv₂²/2 is the kinetic energy at the bottom,

L = Dx(t₁ ) is the work of the drag.

Substituting into (5)

mgx(t₁ ) = mv₂²/2 + Dx(t₁ )

Expressing v₂

v₂ = √(2x(t₁ )×(mg - D)/m) = √(2x(t₁ )×(W - D)×g/W)

Substituting actual numbers

v₂ = √(2×15.6×(4.25 - 0.25)×9.81/4.25) = 16.97 m/s