A 5kg ball is dropped from a height of 10m
a) find the initial potential energy of the ball
B. Find the kinetic energy of the ball just before it reaches ground.
C. Calculate the velocity before it reaches the ground.
I can help you with C:
You have height, acceleration (9.81), and initial velocity (0) so you use:
Vf ^2 = Vi^2 + 2gh
Correction:
A. PE = M*g*h Joules.
a) potential energy=mgh =5*9.8*10=490j b)kinetic energy=1/2mv^2=0 hope it help
B. PE = M*g*h Joules.
B. KE = PE =
C. V^2 = Vo^2 + 2g*h = 0 + 19.6*10 = 196, V = 14 m/s.
To find the answers, we need to use the formulas for potential energy, kinetic energy, and velocity in relation to the height and mass of the ball.
a) The formula for potential energy is given by PE = mgh, where m is the mass (in kg), g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth) and h is the height (in meters).
PE = 5 kg * 9.8 m/s^2 * 10 m = 490 J
So, the initial potential energy of the ball is 490 Joules.
b) The formula for kinetic energy is given by KE = 0.5 * m * v^2, where v is the velocity (in m/s) and KE is the kinetic energy.
Since the ball is dropped, we can assume it starts from rest at the top, so the initial velocity is 0 m/s. At the moment just before it reaches the ground, all of the potential energy will have converted to kinetic energy.
So, the kinetic energy just before it reaches the ground is equal to the initial potential energy:
KE = 490 J
c) To calculate the velocity of the ball just before it reaches the ground, we can use the equation:
KE = 0.5 * m * v^2
Rearranging the equation, we get:
v^2 = (2 * KE) / m
Substituting the known values, we have:
v^2 = (2 * 490 J) / 5 kg
v^2 = 196 m^2/s^2
Taking the square root of both sides, we find:
v = √(196 m^2/s^2)
v ≈ 14 m/s
So, the velocity of the ball just before it reaches the ground is approximately 14 m/s.