A ball of mass 0.5 kg is released from rest at a height of 30 m. How fast is it going when it hits the ground?
how long does it take to fall?
4.9t^2 = 30
now use v = 9.8t to get the speed.
85485
To find the speed of the ball when it hits the ground, we can use the equations of motion and the principle of conservation of energy.
Step 1: Calculate the potential energy at the starting height.
The potential energy (PE) of an object at a certain height can be calculated using the equation:
PE = m * g * h
where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
Given:
m = 0.5 kg
g = 9.8 m/s^2
h = 30 m
PE = 0.5 kg * 9.8 m/s^2 * 30 m
PE = 147 J
The potential energy at the starting height is 147 Joules.
Step 2: Calculate the kinetic energy at the moment of impact.
According to the principle of conservation of energy, the potential energy at the starting height is converted to kinetic energy (KE) at the moment of impact.
The kinetic energy (KE) can be calculated using the equation:
KE = 0.5 * m * v^2
where m is the mass of the object and v is the velocity.
Given:
m = 0.5 kg
KE = 147 J
147 J = 0.5 * 0.5 kg * v^2
Solving for v:
v^2 = (2 * 147 J) / 0.5 kg
v^2 = 588 J/kg
v = √(588 J/kg)
v ≈ 24.25 m/s
The speed of the ball when it hits the ground is approximately 24.25 m/s.
To find the speed of the ball when it hits the ground, we can use the principle of conservation of mechanical energy. The initial potential energy of the ball when it is at a height of 30 m is given by the equation P.E. = mgh, where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
Given:
Mass of the ball (m) = 0.5 kg
Height (h) = 30 m
Acceleration due to gravity (g) = 9.8 m/s^2
Step 1: Calculate the potential energy at the top.
P.E. = mgh
P.E. = (0.5 kg)(9.8 m/s^2)(30 m)
P.E. = 147 J
Step 2: Using the principle of conservation of mechanical energy, the potential energy at the top (P.E.) is converted into kinetic energy at the bottom. Therefore, the kinetic energy when the ball hits the ground will also be 147 J.
Kinetic energy (K.E.) = 147 J
Step 3: Calculate the speed of the ball (v) using the equation for kinetic energy:
K.E. = 1/2 mv^2
Rearranging the equation to solve for v:
v = √(2K.E. / m)
v = √(2 * 147 J / 0.5 kg)
v = √(294 J / 0.5 kg)
v = √588 m^2/s^2 / 0.5 kg
v = √1176 m^2/s^2 / kg
v = 34.28 m/s (rounded to two decimal places)
Therefore, the ball is going approximately 34.28 m/s when it hits the ground.