A brick falls off a scaffold at a height of 80meters above the ground.
Calculate the magnitude of its velocity when it hits the ground.
Air friction is significant. Ignoring that,
PE at top=KE at bottom
mgh=1/2 m v^2
v=sqrt(2*9.8*80)
To calculate the magnitude of the velocity when the brick hits the ground, we can use the equations of motion and the principle of conservation of energy.
First, let's determine the initial velocity of the brick when it falls from a height of 80 meters. Assuming there is no initial vertical velocity and neglecting air resistance, the initial velocity can be considered as zero.
Next, we can use the equation of motion for vertical free-fall:
v^2 = u^2 + 2as
where:
- v is the final velocity (which is what we want to find)
- u is the initial velocity (which is 0 in this case)
- a is the acceleration due to gravity (approximately 9.8 m/s^2)
- s is the displacement (which is -80 meters, considering downward as negative)
Plugging in the values, the equation becomes:
v^2 = 0^2 + 2 * 9.8 * (-80)
Simplifying this equation, we get:
v^2 = -1568
Then, taking the square root of both sides of the equation, we find:
v = √(-1568)
Since we are dealing with the magnitude of velocity, we disregard the negative sign. Thus, the magnitude of the velocity when the brick hits the ground is:
v = √1568 m/s, which is approximately 39.6 m/s.