Question

3. A ball of mass m=0.5kg is thrown at 1 meter above the ground (meaning X; = 0, Y; = 1m) and with an initial velocity of Vo=600 m/s along degree (with respect to the horizontal X-axis). a) Starting from Newton's second law, find V (t), Vy(t), X(t), and Y(t). b) From a), find the range R and the maximum height Y max the ball can reach. c) i) Plot the trajectories of the ball using the results in a) for = 41, 43, 45, 47, 49 degrees, respectively (plotting them on the same graph, taking a whole page). ii) Find the range R and maximum height Ymax for each angle from the plot. d) Now, we add the effect of air resistance, which is given by Fair=- mkV (k=0.05/sec), to the motion of the ball. Answer the same questions in a) and c) e) Discuss the effects on R and Ymax due to the air resistance by citing the results from c) and d).

Answer 1

(a) Horizontal velocity component is unaffected by gravity , Hence we have

        V_x(t) = V_o cos$\theta$   ..........................(1)

Vertical velocity component is obtained from equation of motion v = u - a t , where u is initial velocity, v is velocity after time t and a is retardation . In this case a = g , where g is acceleration due to gravity

Hence , V_y(t) = ( V_o sin$\theta$ ) - g t ......................(2)

Horizontal displacement, X(t) = Horizontal velocity $\times$ time

X(t) = ( V_o cos$\theta$ ) t ........................(3)

Vertical displacement is obtained from equation of motion S = ut - (1/2)at²

Y(t) = ( V_o sin$\theta$ ) t - (1/2) g t² .....................(4)

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(b) if t_u is time taken by the ball to reach maximum height, t_u is obtained from eqn.(2) by substituting V_y( t_u ) = 0

t_u = ( V_o sin$\theta$ ) / g ...................... (5)

if T is time taken for the mass to reach same horizontal position Y = 1 m, when it is falling down,

then T = 2 t_u = 2 ( V_o sin$\theta$ ) / g .......................(6)

During this time T, horizontal distance R₁ traveled is given by,

R₁ = V_x(t) $\times$ T = V_o cos$\theta$$\times$ 2 ( V_o sin$\theta$ ) / g = V_o² sin2$\theta$ / g    ...................(7)

After reaching the position Y= 1m, ball will travel 1m down and time taken t to hit the ground level Y = 0

is obtained from the following equation

( V_o sin$\theta$ )t + (1/2)gt² = 1

solving for t from above eqn. , we get time t as

+-v, + 402 + 2g

Horizontal distance R₂ traveled during this time t is given by

R 402 +29 x .coso = 0 + 402 + 20

Range = R₁ + R₂

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Maximum height reached is obtained from equation of motion, " v² = u² - 2 a S " by substituting final speed v = 0

Hence maximum height H = ( v_o² sin²$\theta$ ) / (2g)

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With drag due to air resistance

Horizontal retardation,

I = -mku. dt

1 dt, X . = -dt

By integrating both sides, we get

.............(8)

1 x In(U2) = - +

at t = 0, we have v_x = v_o cos$\theta$ . Hence C = ln (v_o cos$\theta$ )

Hence eqn.(8) becomes,

vocos
   or   
Vr(t) = v.coste-mkt

Horizontal displacement with drag

dX(t) dt -= v.cosde-mkt

By integration, we get

X(t) - Vocoste-mkt - + -mk

at t = 0, X = 0 , Hence C = ( v₀ cos$\theta$ ) / ( m k )

Hence we get

e-mkt X(t) = Pocos mk1

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Similarly we write acceleration for vertical movement

y = -9 - mkv

dug =-dt 9 + mku

by integrating we get

Vy = (-9. + vosino) e-mkt - 9 mk