V^2 = Vo^2 + 2g*h = 0, h = -Vo^2/2g = -80/-19.6 = 4.08 m. =
max. ht.
1. PE = M*g*h/2 = 0.1*9.8*(4.08/2) = 2.0 Joules.
2. PE = 0.1*9.8*4.08 = 4.0 Joules.
3. KE = 0.5M*V^2 = 0.5*0.1*80^2 = 320 Joules.
a ball of mass 0.1kg is thrown vertically upwards with an initial velocity of 80m/s.calculate the potential energy (I) half way up (ii) at its maximum height.what is the kinetic energy as it leaves the ground
4 answers
at max ht v=0
therfore u^2=2gh
h=u^2/2g
for halway do h\2
therfore u^2=2gh
h=u^2/2g
for halway do h\2
time
hw did u get u max height? i did not understand
1 answer