Respuesta :
Answer:
a) v = 21.34 m/s
b) v = 21.34 m/s
c) v = 21.34 m/s
Explanation:
Mass of the snowball, m = 0.560 kg
Height of the cliff, h = 14.2 m
Initial velocity of the ball, u = 13.3 m/s
θ = 26°
The speed of the slow ball as it reaches the ground, v = ?
The initial Kinetic energy of the snow ball, [tex]KE_{0} = 0.5 mu^{2}[/tex]
Potential energy of the snow ball at the given height, PE = mgh
Final Kinetic energy of the ball as it reaches the ground, [tex]KE_{f} = 0.5mv^{2}[/tex]
a) Using the principle of energy conservation,
[tex]KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s[/tex]
b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch
c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass
v = 21. 34 m/s
