Posted by cyrus on Saturday, November 8, 2008 at 10:50pm.
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.
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.
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