a.KE = 0.5m*V^2 = 0.5*1.4*8.5^2=50.6 J.
b. V^2 = Vo^2 + 2g*d
V^2 = 8.5^2 + 19.6*37 = 725.2
V = 26.9 m/s.
c. 0.5m*V^2 = 3600 J.
.5*1.4*V^2 = 3600
0.7V^2 = 3600
V^2 = 5143
V = 71.7 m/s. When it hits the gnd.
V^2 = Vo^2 + 2g*h
Vo^2 = V^2-2g*h
Vo^2 = (71.7)^2 - 19.6*37 = 4416
Vo = 66.5 m/s. Gives the rock 3.6 kJ of
energy.
A hiker throws a rock with a mass of 1.4kg from a cliff that is 37 m above the ground . The hiker gives the rock an initial downward velocity of 8.5 m/s.
a) What is the energy of the rock at the moment the hiker releases it?
b) What is the velocity of the rock at the moment it hits the ground?
c) What velocity would the hiker have had to give the rock in order for it to have 3.6 kJ of energy when it hits the ground?
1 answer
2 answers