You are in an egg drop contest which you win by breaking your egg the hardest. You drop your egg off of a 3 story building (12m above the ground). You can throw it up with speed 11m/s, or straight down with speed 11m/s.
How high above the ground will the upward thrown egg reach in m?
To solve this problem, we can use the kinematic equation:
h = h0 + v0t - 1/2gt^2
where:
h = final height (what we're solving for)
h0 = initial height (12m above ground)
v0 = initial velocity (11m/s upwards)
g = acceleration due to gravity (-9.8m/s^2)
t = time (what we're solving for)
We want to find the height reached by the egg when thrown upwards, so we'll set the vertical velocity to 0 at the maximum height (when the egg is at the top of its trajectory). This gives us:
0 = v0 - gt
t = v0/g
Plugging this in to the first equation, we get:
h = h0 + v0(v0/g) - 1/2g(v0/g)^2
h = 12 + 11(11/-9.8) - 1/2(-9.8)(11/-9.8)^2
h = 12 - 60.5 + 6.06
h = -42.44
Uh-oh! This means that the egg does not reach a positive height when thrown upwards, so it would break upon impact with the ground. It turns out that the initial velocity is not high enough to overcome the acceleration due to gravity and reach a positive height.
Therefore, the best strategy would be to drop the egg straight down from the building (with an initial velocity of 0) in order to break it the hardest.
Well, if I were to throw the egg directly upwards with a speed of 11m/s, it would definitely reach a certain height before gravity inevitably brings it back down. To find out how high it reaches, we can use some physics!
The equation we need to use is the kinematic equation for vertical motion:
v² = u² + 2as
Where:
v = final velocity (0m/s since the egg eventually stops)
u = initial velocity (11m/s)
a = acceleration due to gravity (-9.8m/s², because gravity always brings everything down)
s = displacement (the height we want to find)
Since we're trying to find the upward distance the egg goes before stopping, we're looking for s. We can rearrange the equation to solve for it:
s = (v² - u²) / (2a)
Substituting the values into the equation, we get:
s = (0 - 11²) / (2 * -9.8)
Calculating that out, we find:
s ≈ -11.27m
Now, since distance can't be negative, my upward thrown egg will reach a height of approximately 11.27 meters above the ground! Just make sure to catch it on the way down, or you might have an egg-cellent mess to clean up!
To determine how high the upward thrown egg will reach, we can use the equations of motion. The initial speed of the egg thrown upward is 11m/s.
The equation we can use is:
v^2 = u^2 - 2as,
where v is the final velocity (0m/s when the egg reaches its highest point), u is the initial velocity (11m/s), a is the acceleration due to gravity (-9.8m/s^2), and s is the displacement or the height we are trying to find.
Plugging in the given values into the equation, we have:
0^2 = 11^2 - 2(-9.8)s.
Simplifying the equation, we get:
0 = 121 + 19.6s.
Rearranging the equation to solve for s, we get:
19.6s = -121.
Dividing both sides of the equation by 19.6, we have:
s = -121 / 19.6.
Therefore, the upward thrown egg will reach a height of approximately 6.17m above the ground.
To determine how high above the ground the upward thrown egg will reach, we can use the principles of projectile motion. We can start by identifying the known values:
Initial velocity (u) = 11 m/s (upward)
Acceleration due to gravity (a) = -9.8 m/s² (downward)
Height of the building (h) = 12 m
The equation we can use is:
v² = u² + 2aΔy
Where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity (11 m/s)
a = acceleration due to gravity (-9.8 m/s²)
Δy = change in vertical position (maximum height above the ground)
Since the final velocity is zero at the maximum height, we can rewrite the equation as:
0 = (11 m/s)² + 2(-9.8 m/s²)Δy
Simplifying and solving for Δy:
0 = 121 m²/s² - 19.6 m/s² Δy
19.6 m/s² Δy = 121 m²/s²
Δy = 121 m²/s² / 19.6 m/s²
Δy ≈ 6.17 m
Therefore, the upward thrown egg will reach a height of approximately 6.17 meters above the ground.