Question

A projectile of mass m = 5kg is thrown upward vertically with a velocity v₀ = 9 m/s

a) Starting with Newton's 2nd Law F_Net = ma, calculate the time to an accuracy of four decimal places at which the maximum height is reached when no air resistance is present.

b) Starting with Newton's 2nd Law F_Net = ma, calculate the time to an accuracy of four decimal places at which the maximum height is reached when considering the following force of air resistance:

F_air = 7.775 x 10^-5v

c) Starting with Newton's 2nd Law F_Net = ma, calculate the time to an accuracy of four decimal places at which the maximum height is reached when considering the following force of air resistance:

F_air = 0.055v²

Answer 1

mass m = 5 kg

initial velocity   = 9 m/s

The object moves upward weight of the body mg pulls it down

force acting on the body F_net = mg

acceleration of the body = F/m = g = 9.8 m/s/s - acting opposite to the motion of the object

it moves up until its velocity becomes 0 - final velocity

v_f = v_o -gt

t = v_o/g = 9/9.8 = 0.9184 s

b) air resistance F_air = 7.775E-5 v

F_net = ma = mg + rv ( r= 7.775E-5)

acceleration a = dv/dt = -g- r/m v

dv/(g+r/m v) = dt

integrating on both sides we get

m/r ln(g + r/m v) = -t + C , C - integration const.

at t= 0 , v = = 9 m/s

C = m/r ln(g + r/m )

final velocity v =0 at max -height

t = C - m/r ln(g ) = m/r * ln( 1 + r/mg) = 0.9184 s

c) F_air = 0.055 v²

dv/dt = -g - 0.011v²

dv/(g+0.011v² ) = -dt

dv/(890.9 + v² ) = - 0.011 dt

1/sqrt(890.9) Tan^-1 ( v/sqrt(890.9) ) = -0.011 t +C

at t=0 ; v = 9.0

C = 0.0098

at max height v=0

-0.011 t +C =0

t = 0.0098/0.011 = 0.8920 s