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?
1
Expert's answer
2010-11-10T02:44:07-0500
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 = v2/2g; S1/2 = v2/4g = 4.56 m. v2/4g = vt - gt2/2; 4.56 = 13.5t - 5t2; t1 = 0.4 s; t2 = 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 = mv2/2 = 0.345* 9.52 / 2 = 6.2 J
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