An egg is thrown from the top edge of a 22.9 m high building down onto an unsuspecting person standing on the ground 6.2 m from the base of the building. If the egg is tossed horizontally, at what speed, in m/s, must the egg be thrown such that it lands on the person?
time to fall:
h=1/2 g t^2
t=sqrt(2*22.9/9.8)
horizontal
v=d/t=6.2/t
Why did the egg go to school? Because it wanted to get a "crack" education!
To solve this egg-cellent problem, we can use some physics! Since the egg is thrown horizontally, its initial vertical velocity is 0 m/s. The only force acting on it is gravity, causing it to accelerate downward at 9.8 m/s².
Now let's find the time it takes for the egg to reach the ground. We can use the formula:
h = (1/2) * g * t²
Where:
h = height of the building (22.9 m)
g = acceleration due to gravity (9.8 m/s²)
t = time
Rearranging the equation, we get:
t = sqrt((2 * h) / g)
t = sqrt((2 * 22.9) / 9.8) ≈ 2.42 seconds (rounded to 2 decimal places)
Next, we can find the horizontal distance the egg travels during this time. Using the formula:
d = v * t
Where:
d = horizontal distance (6.2 m)
v = initial horizontal velocity (what we're trying to find, in m/s)
t = time (2.42 s)
Rearranging the equation, we get:
v = d / t
v = 6.2 / 2.42 ≈ 2.56 m/s (rounded to 2 decimal places)
So, the egg must be thrown horizontally at a speed of approximately 2.56 m/s to reach the person standing at 6.2 m from the base of the building. Just make sure the eggs-periment doesn't end up eggs-ploding!
To determine the required speed at which the egg must be thrown horizontally, we can set up a basic projectile motion problem.
Given:
- Initial vertical height (h) = 22.9 m
- Horizontal distance (d) = 6.2 m
- Acceleration due to gravity (g) = 9.8 m/s^2
We need to find the horizontal velocity (v) at which the egg must be thrown.
Since there is no initial vertical velocity, the time for the egg to fall from the top of the building to the ground can be found using the equation:
h = (1/2) * g * t^2
Rearranging the equation:
t^2 = (2 * h) / g
Substituting the given values:
t^2 = (2 * 22.9) / 9.8
t^2 = 4.6735
t ≈ √4.6735
t ≈ 2.16 s (approx.)
Now, we can find the horizontal velocity (v) using the equation:
v = d / t
Substituting the given values:
v = 6.2 / 2.16
v ≈ 2.87 m/s (approx.)
Therefore, the egg must be thrown horizontally at a speed of approximately 2.87 m/s to land on the person.
To determine the required speed at which the egg must be thrown horizontally to hit the person, we can use the principles of projectile motion.
When the egg is thrown horizontally, it will have an initial horizontal velocity (Vx) and an initial vertical velocity (Vy) of 0. The only force acting on the egg in the horizontal direction is the constant velocity in that direction. In the vertical direction, the only force acting on the egg is the acceleration due to gravity (g = 9.8 m/s^2).
To find the time (t) it takes for the egg to fall from the top of the building to the ground, we can use the equation:
h = (1/2) * g * t^2
where h is the vertical distance traveled (22.9 m in this case) and t is the time.
Solving for t:
22.9 = (1/2) * 9.8 * t^2
t^2 = (22.9 * 2) / 9.8
t^2 = 4.663
t ≈ √4.663
t ≈ 2.16 seconds
Now that we have the time it takes for the egg to fall, we can calculate the horizontal distance traveled (d) using the equation:
d = Vx * t
where d is the horizontal distance traveled (6.2 m in this case), Vx is the horizontal velocity of the egg, and t is the time.
Solving for Vx:
6.2 = Vx * 2.16
Vx = 6.2 / 2.16
Vx ≈ 2.87 m/s
So, the egg must be thrown horizontally with a speed of approximately 2.87 m/s in order to land on the person standing 6.2 m from the base of the building.