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December 21, 2014

December 21, 2014

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

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.

- math/physics -
**drwls**, Saturday, November 8, 2008 at 11:12pmThis 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.

- math/physics -
**cyrus**, Saturday, November 8, 2008 at 11:15pmhow would you solve it with differential equations

- math/physics -
- math/physics -
**drwls**, Saturday, November 8, 2008 at 11:32pmSolve

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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