an egg is thrown directly upward so that it remains in the air for 10 seconds.what is the initial velocity of the egg?
well, let's do the first half of the problem, upward, symmetrical velocity profile during fall
g = -9.81 m/s^2
v = Vi - g t
v = Vi - 9.81 t
v = 0 at top in 5 seconds
0 = Vi - 9.81(5)
Vi = 49 m/s
To determine the initial velocity of the egg, we can use the equation of motion for vertical motion. The equation is:
h = (v₀ * t) - (0.5 * g * t²)
Where:
h = height (in this case, it is 0, since the egg starts and ends at the same height)
v₀ = initial velocity
t = time (10 seconds in this case)
g = acceleration due to gravity (approximately 9.8 m/s²)
Plugging in the values, we get:
0 = (v₀ * 10) - (0.5 * 9.8 * 10²)
Rearranging the equation:
(0.5 * 9.8 * 10²) = (v₀ * 10)
Simplifying further:
v₀ = (0.5 * 9.8 * 10²) / 10
v₀ = (0.5 * 9.8 * 100) / 10
v₀ = (4.9 * 100) / 10
v₀ = 49 m/s
Therefore, the initial velocity of the egg is 49 m/s.