A Q. 200 kg ball is thrown upward with an initial speed of 24.0 m/s at an angle

Question

A Q. 200 kg ball is thrown upward with an initial speed of 24.0 m/s at an angle of 30.0 degree with the horizontal from the edge of a 60.0 m cliff. Neglecting air resistance, find (a) the initial kinetic energy of the ball, (b) the potential energy and kinetic energy of the ball when it is at its highest point, (c) the kinetic energy of the ball when it reaches the ground 60 m below the cliff, and (d) the speed of the ball just before it strikes the ground. Assume that U = 0 at the top of the cliff.

Explanation / Answer

part a:

initial kinetic energy=0.5*mass*initial speed^2

=0.5*0.2*24^2=57.6 J

initial potential energy=0

total initial mechanical energy=initial potential energy+initial kinetic energy

=57.6 J

part b:

at the highest point, speed=0==>kinetic energy =0

as total energy is conserved, potential energy=initial total mechanical energy

=57.6 J

part c:

when the ball reaches ground, potential energy=mass*g*height

=0.2*9.8*(-60)=-117.6 J

then kinetic energy+potential energy=total mechanical energy

==>kinetic energy-117.6=57.6

==>kinetic energy=175.2 J

part d:

speed of the ball just before striking the ground=sqrt(2*kinetic energy/mass)

=sqrt(2*175.2/0.2)=41.86 m/s

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