Mr.Notten throws an egg from the top of a school at an initial velocity of 20m/straight upward. The egg is launched 50 m above the ground, and the egg just misses the edge of the roof on its way down? what is maximum height? what is the maximum height of the egg? Determine the velocity of the egg when it returns to the height from which it was thrown.
To find the maximum height reached by the egg, we can use the kinematic equation for the upward motion:
Vf^2 = Vi^2 + 2aΔx
Where:
- Vf is the final velocity (which is 0 when the egg reaches its highest point).
- Vi is the initial velocity (20 m/s, as given).
- a is the acceleration due to gravity, which is -9.8 m/s^2 (negative since it's in the opposite direction).
- Δx is the displacement in the vertical direction (50 m above the ground).
Using the equation:
0^2 = (20^2) + 2(-9.8)(Δx)
0 = 400 - 19.6(Δx)
19.6(Δx) = 400
Δx = 400/19.6 ≈ 20.41 m
Therefore, the maximum height reached by the egg is approximately 20.41 m.
To find the velocity when the egg returns to the height from which it was thrown, we can use the kinematic equation for the downward motion:
Vf = Vi + at
Where:
- Vf is the final velocity (which is what we're looking for).
- Vi is the initial velocity (which is 0 when the egg reaches its highest point).
- a is the acceleration due to gravity, which is -9.8 m/s^2.
- t is the time taken for the egg to reach the maximum height.
Since the upward and downward motions follow the same path, the time taken to reach the maximum height is the same as the time taken to return to the height from which the egg was thrown.
Using the equation:
Vf = 0 + (-9.8)t
To determine t, we can use the equation:
Δx = Vit + (1/2)at^2
Where:
- Δx is the displacement (50 m above the ground).
- Vi is the initial velocity (20 m/s).
- a is the acceleration due to gravity (-9.8 m/s^2).
- t is the time taken.
Using the equation:
50 = (20)t + (1/2)(-9.8)t^2
50 = 20t - 4.9t^2
Rearranging the equation:
4.9t^2 - 20t + 50 = 0
Solving this quadratic equation, we find two possible solutions for t, but we know that the one where the egg is on its way down is the correct one. Let's take that solution.
t ≈ 2.04 seconds
Now, using this value of t in the equation for Vf:
Vf = 0 + (-9.8)(2.04)
Vf ≈ -19.92 m/s
Therefore, the velocity of the egg when it returns to the height from which it was thrown is approximately -19.92 m/s, which means it is moving downward.
To determine the maximum height reached by the egg, we can use the equations of motion under constant acceleration. In this case, we can consider the upward motion of the egg as positive.
First, let's identify the given values:
Initial velocity (u) = +20 m/s (since it is thrown upward)
Initial height (h) = +50 m (above the ground)
Acceleration due to gravity (g) = -9.8 m/s² (negative because it acts downward)
1. Calculate the time taken to reach the maximum height:
We can use the equation: v = u + gt, where v represents the final velocity.
Since the egg reaches the maximum height, the final velocity at that point would be zero (v = 0).
Rearranging the equation, we have: t = (v - u) / g.
t = (0 - 20) / -9.8 = 2.04 seconds
2. Calculate the maximum height (H) using the formula: H = h + ut + (1/2)gt².
Substituting the given values:
H = 50 + (20)(2.04) + (1/2)(-9.8)(2.04)² = 61.632 m.
Therefore, the maximum height reached by the egg is 61.632 meters above the ground.
3. Determine the velocity when the egg returns to the initial height:
Using the equation: v = u + gt, we can calculate the final velocity at the initial height.
t = 2.04 seconds (calculated earlier)
v = ?
u = +20 m/s
g = -9.8 m/s²
v = 20 + (-9.8)(2.04) = 0.392 m/s (rounded to three decimal places).
Hence, the velocity of the egg when it returns to the height from which it was thrown is approximately 0.392 m/s.