I built a structure to protect an egg as we dropped it from increasing heights. Its last height was 3 m and it weighted .093 lbs. Can someone tell me how to calculate the final velocity? I don't know how to do this by ignoring time.
weighed*
Fell 3 meters, ignoring air friction:
V = a T = 9.8 T
so T = V/9.8 where T is the time in the air
then distance
h= (1/2) a t^2
h = .5 (9.8)(V/9.8)^2
3 = .5 V^2/9.8
6*9.8 = V^2
7.67 m/s
To calculate the final velocity of an object dropped from a height, you can use the principle of conservation of mechanical energy. The potential energy of the object at the starting height is converted into kinetic energy at the final height. The formula to calculate the final velocity is derived from the conservation of mechanical energy:
1/2 * m * v^2 = m * g * h
Where:
m is the mass of the object
v is the final velocity
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height from which the object is dropped
In this case, we have the height of 3 m and the weight of 0.093 lbs. Before we proceed with the calculation, we need to convert the weight from pounds to kilograms, as the standard unit for mass in the formula is kilograms.
1 lb ≈ 0.4536 kg
Converting the weight of the object:
0.093 lbs * 0.4536 kg/lb ≈ 0.0422 kg
Now, we can substitute the known values into the formula to find the final velocity:
1/2 * 0.0422 kg * v^2 = 0.0422 kg * 9.8 m/s^2 * 3 m
Simplifying the equation:
0.0211 kg * v^2 = 0.1242 kg * m^2/s^2
Dividing both sides of the equation by 0.0211 kg:
v^2 ≈ 5.8778 m^2/s^2
Taking the square root of both sides:
v ≈ √(5.8778) ≈ 2.4259 m/s
Therefore, the final velocity of the object when it hits the ground (from a height of 3 m) is approximately 2.43 m/s.