An egg is dropped from 75 cm, what velocity is egg at when it hits the floor.
What is the maximum height it can be dropped from to withstand impact velocity of 1.7ms-1
I used
V*2 = v*2 + 2ad
V = sqrt 0 + 2 x -9.8 x .75 = -3.83m/s
0.14744 m rounded to 0.15 m
yes,
good to remember v = -sqrt (2 g h)
here
v = -sqrt (2*9.81*.75)
= -3.84 m/s
1.7 = sqrt (2 *9.81 * h)
h = .1472986748
I think g is 9.81, you use 9.8 some classes use 10
Thanks again !
To find the velocity of the egg when it hits the floor after being dropped from a height of 75 cm, you correctly used the equation for motion under gravity:
V^2 = u^2 + 2ad
Where V is the final velocity, u is the initial velocity, a is acceleration (in this case due to gravity, which is approximately -9.8 m/s^2), and d is the distance traveled.
Plugging in the values:
V^2 = 0^2 + 2 * (-9.8) * 0.75
V^2 = -19.6 * 0.75
V^2 = -14.7 m^2/s^2
Since we are interested in the velocity at impact, V should be positive (as speed is always a positive quantity). Therefore, we take the square root of V^2 to find the magnitude of the velocity:
V = √(-14.7) m/s
However, it's worth noting that we encounter a problem—taking the square root of a negative number results in an imaginary answer, which doesn't have any physical significance in this case.
So, it seems there may be an error in the calculations. When working with the given problem, make sure to double-check all the steps of the calculation to identify any possible mistakes.
Moving on to the second part of your question about the maximum height the egg can be dropped from to withstand an impact velocity of 1.7 m/s:
To determine this, we can rearrange the same equation:
V^2 = u^2 + 2ad
Rearranging for u (initial velocity), we get:
u^2 = V^2 - 2ad
Substituting the given values:
u^2 = (1.7)^2 - 2 * (-9.8) * d
Simplifying:
u^2 = 2.89 + 19.6d
We want to find the height at which the initial velocity is zero, so we set u = 0:
0 = 2.89 + 19.6d
Solving for d:
19.6d = -2.89
d = -2.89 / 19.6
d ≈ -0.1474 m
Since height cannot be negative, we discard the negative value. Therefore, the maximum height the egg can be dropped from to withstand an impact velocity of 1.7 m/s is approximately 0.1474 meters, rounded to 0.15 meters.
It's important to note that in this calculation, we have assumed no air resistance and perfect elasticity between the egg and the floor. In reality, there may be other factors that affect the result.