A ball of mass m is thrown straight up into the air with an initial speed v0.
(a) Find an expression for the maximum height reached by the ball in terms of v0 and g.
(b) Using conservation of energy and the result of part (a), find the magnitude of the momentum of the ball at one-half its maximum height in terms of m and v0.
Thank you for the opportunity to help you with your question!
a)H maximum = Hmax at maximum height V = 0 then gt = Vo ==> t = Vo/g Hmax = (Vo)²/g - (Vo)²/(2g) Hmax = (Vo)²/(2g)
b) conservation of energy means that energy, KE+PE, is a constant.
m(Vo)²/2 + 0 = mgH + mV²/2 V² = (Vo)² - 2gH Hmax/2 = (Vo)²/(4g) V² = (Vo)² - 2g(Vo)²/(4g) V² = (Vo)² - (Vo)²/2 V² = (Vo)²/2 ==> V = ((Vo)√2)/ 2 mV = (m(Vo)√2)/2
NOTE: g is the gravity=9.8N
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?