Question #971

You throw a 0.345kg ball straigt up with an initial velocity of 13.5m/s
a. what is the momentum of the ball at the highest point of its path?
b. what is its momentum halfway up?
c. what is its kinetic energy halfway up?

Expert's answer

a. At the highest point of the motion the velocity equals to zero, p = mv = m*0 = 0;b. From the equations of the motion S = v^{2}/2g; S_{1/2} = v^{2}/4g = 4.56 m.

v^{2}/4g = vt - gt^{2}/2;

4.56 = 13.5t - 5t^{2};

t_{1} = 0.4 s; t_{2} = 2.3 s.

v(t=0.4) = v-gt = 13.5 - 0.4*10 = 9.5 m/s.

The momentum is p= 0.345* 9.5 = 3.3 kg* m/s;

The energy is E = mv^{2}/2 = 0.345* 9.5^{2} / 2 = 6.2 J

v

4.56 = 13.5t - 5t

t

v(t=0.4) = v-gt = 13.5 - 0.4*10 = 9.5 m/s.

The momentum is p= 0.345* 9.5 = 3.3 kg* m/s;

The energy is E = mv

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