A baseball of mass m = 0.145 kg is dropped from the top of a tall building 302.0

Question

A baseball of mass m = 0.145 kg is dropped from the top of a tall building 302.0 m above street level. The baseball
lands on the center of a trampoline whose surface (when
at and at rest) is 2.00 m above street level. The trampoline
can be treated as a spring with spring constant k = 150.0 N/m, where the displacement of the center of the trampoline
corresponds to the displacement x of the spring. The coecient of air resistance for the baseball is
a= 1.30 x 10^-3 Ns^2/m^2.

Assuming that no frictional forces act while the ball stretches the trampoline, what will be the maximum distance
that the center of the trampoline is stretched downward by the baseball?

Explanation / Answer

energy of ball just before it hits trampoline = potential energy + kinetic energy
= mgh+ mv2/2

velocity of ball just before it hits trampoline = (2 x(302-2))/(9.81-1.3 x 10-3)

=76.71 m/s

energy of ball just before it hits trampoline = 0.145 x 9.81 x 2 + 0.145 x 76.712 /2

=429.46 J

applying law of conservartion of energy

energy of ball just before it hits trampoline = energy at streched position

= kx2/2 + mgx ( x is maximum distance stretched)

429.46 = 150x2/2 + 0.142 x

solving this quadratic equation we get x = 1.25m

maximum distance stretched = 1.25m

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