7)
An apple falls from a tree (h = 9m). In the absence of air resistance, how fast is it traveling when it strikes the ground?
vf^2=vi^2 +2ad
a=9.8 m/s^2
solve for vf
13.3
To calculate the speed at which the apple strikes the ground, we can use the principles of physics.
First, we need to determine the acceleration due to gravity, which is approximately 9.8 m/s² near the Earth's surface.
Next, we can use the formula for velocity:
v = √(2gh)
Where:
v = final velocity (speed at which the apple strikes the ground)
g = acceleration due to gravity (approximately 9.8 m/s²)
h = height from which the apple falls (9 m)
Plugging in the values, we get:
v = √(2 * 9.8 * 9)
Calculating the value inside the square root:
v = √(176.4)
v ≈ 13.3 m/s
Therefore, the apple is traveling at approximately 13.3 m/s when it strikes the ground.
To calculate the speed at which the apple strikes the ground, you can use the principle of conservation of energy. The potential energy lost by the apple as it falls is equal to the kinetic energy gained.
First, calculate the potential energy (PE) lost by the apple:
PE = m * g * h
Where:
m = mass of the apple
g = acceleration due to gravity (approximately 9.8 m/s²)
h = height of the tree (9m)
Next, calculate the kinetic energy (KE) gained by the apple:
KE = 1/2 * m * v²
Where:
v = velocity of the apple
According to the principle of conservation of energy, the potential energy lost is equal to the kinetic energy gained:
PE = KE
Therefore, we can equate the two equations and solve for the velocity (v):
m * g * h = 1/2 * m * v²
Simplify the equation by canceling out the mass (m) on both sides:
g * h = 1/2 * v²
Now, isolate the velocity (v) by multiplying both sides of the equation by 2 and taking the square root:
2 * g * h = v²
v = sqrt(2 * g * h)
Substituting the given values:
v = sqrt(2 * 9.8 m/s² * 9 m)
Calculating the expression:
v ≈ sqrt(176.4) ≈ 13.3 m/s
Therefore, the apple will be traveling at approximately 13.3 m/s when it strikes the ground.