Question
a ball with mass 0.15 kg is thrown upward with initial velocity 20 m/sec from the roof of a building 30 m high. there is a force due to air resistance of |v|/30, where velocity v is measured in m/sec.
a. find the maximum height above the ground the ball reaches.
b. find the time the ball hits the ground.
you cannot use the kinematic equations.
a. find the maximum height above the ground the ball reaches.
b. find the time the ball hits the ground.
you cannot use the kinematic equations.
Answers
This IS a kinematics problem. Any equation used to solve it involves kinematics. An exact solution would require solving a (kinematic) differential equation.
A good approximation to part (a) can be obtained by assuming an average air resistance force of |V|/60 , where V is the initial velocity of 20 m/s. Thus work done against friction going up is
V/60*H, and initial kientic energy equals potential energy gain at the highest elevation PLUS work done against friction.
(1/2) M V^2 = M g H + V/60*H
H = (1/2)V^2/[g + V/(60M)]= (1/2)V^2/11
= 18.2 m
The time spent by the ball going up is
very nearly H/(V/2) = 0.46 s
Use a similar energy conservation argument to estimate the (longer) time it takes for the ball to come back down. Note that it must travel an additional 30 m to reach the ground.
A good approximation to part (a) can be obtained by assuming an average air resistance force of |V|/60 , where V is the initial velocity of 20 m/s. Thus work done against friction going up is
V/60*H, and initial kientic energy equals potential energy gain at the highest elevation PLUS work done against friction.
(1/2) M V^2 = M g H + V/60*H
H = (1/2)V^2/[g + V/(60M)]= (1/2)V^2/11
= 18.2 m
The time spent by the ball going up is
very nearly H/(V/2) = 0.46 s
Use a similar energy conservation argument to estimate the (longer) time it takes for the ball to come back down. Note that it must travel an additional 30 m to reach the ground.
how would you solve it with differential equations
Solve
M dv/dt = -M g -v/60*H
(v>0)
followed by
M dv/dt = -M g + v/60*H
(v<0)
Once you have v(t), integrate v dt to get the distance travelled vs t.
M dv/dt = -M g -v/60*H
(v>0)
followed by
M dv/dt = -M g + v/60*H
(v<0)
Once you have v(t), integrate v dt to get the distance travelled vs t.