Question

A ball of mass m=2.00 kg is thrown vertically with velocity v0=10.0 m/s. a) What maximum height would it reach if the average force of air resistance were Fd= 5 N? [4.06 m] b) The ball lands on a spring (with k=200 N/m), which brings it to a stop 1 m above the launch point. How much would the spring compress before the ball stops? (Assume that air resistance acts the entire way) [0.67 m] I don't know how to approach part b. Thank you!

Answer 1

Net Acceleration acting on the mass during upward motion

a = - (5/2 + 9.8) = - 12.3 m/s^2

Using 3rd equation of motion

v^2 = u^2 + 2 ad

d = u^2/(2a)

d = 10^2/(2*12.3)

d = 4.065 m

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b)

Considering energy equilibrium

Total energy before throw = total energy after stop

0.5mu^2 - mgh = energy dissipation by air friction + spring potential energy

0.5 * 2*10^2 - 2*9.8* 1 = ( 5*4.065 + 5* 3.065) + 0.5 *200 x^2

x = 0.67 m

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