An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point a distance 48.0 m below its starting point at a time 5.00s after it leaves the thrower's hand. Air resistance may be ignored.
part A:
What is the initial speed of the egg?
v= m/s
part B:
How high does it rise above its starting point?
h= m
part C:
What is the magnitude of its velocity at the highest point?
v= m/s
part D:
What is the magnitude of its acceleration at the highest point?
a= m/s2
part E:
What is the direction of its acceleration at the highest point?
up or down
part a and part b
ege
To solve this problem, we can use kinematic equations. Let's start solving each part step by step:
Part A:
To find the initial speed (v₀), we need to use the equation:
v = v₀ - gt
Where:
v = final velocity (at the time of 5.00s after the throw)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time (5.00s)
Since the egg is thrown nearly vertically upward and just misses the cornice, the final velocity (v) is 0 m/s at the highest point. Plugging these values in, we can solve for the initial speed (v₀):
0 = v₀ - (9.8 m/s^2)(5.00s)
Rearranging the equation, we have:
v₀ = (9.8 m/s^2)(5.00s)
v₀ = 49.0 m/s (rounded to one decimal place)
Therefore, the initial speed of the egg is 49.0 m/s.
Part B:
To find the maximum height (h), we can use the formula:
h = v₀t - (1/2)gt^2
Where:
h = maximum height
v₀ = initial speed (49.0 m/s from Part A)
t = time (5.00s)
g = acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the values, we have:
h = (49.0 m/s)(5.00s) - (1/2)(9.8 m/s^2)(5.00s)^2
Simplifying the equation:
h = 245 m - 122.5 m
h = 122.5 m
Therefore, the egg rises to a height of 122.5 m above its starting point.
Part C:
At the highest point, the egg momentarily comes to rest, which means its velocity is 0 m/s. Therefore, the magnitude of its velocity at the highest point is 0 m/s.
Part D:
Since the egg momentarily comes to rest at the highest point, its acceleration is equal to the acceleration due to gravity (9.8 m/s^2). Therefore, the magnitude of its acceleration at the highest point is 9.8 m/s^2.
Part E:
The direction of the acceleration at the highest point is downward.
To solve this problem, we can use the kinematic equations of motion. Let's go through each part step by step.
Part A: What is the initial speed of the egg?
To find the initial speed of the egg, we can use the fact that the egg passes a point 48.0 m below its starting point after 5.00 seconds.
We'll start by using the equation of motion, which relates the vertical displacement (Δy), initial velocity (v0), time (t), and acceleration due to gravity (g):
Δy = v0 * t - (1/2) * g * t^2
Since the egg starts from the ground, the initial vertical displacement (Δy) is zero. The equation becomes:
0 = v0 * t - (1/2) * g * t^2
We know that Δy = -48.0 m (negative because the egg is falling below its starting point) and t = 5.00 s. The acceleration due to gravity is approximately -9.8 m/s^2 (negative because it acts in the opposite direction to the motion).
Now we can substitute the known values into the equation and solve for v0:
-48.0 = v0 * 5.00 - (1/2) * (-9.8) * (5.00)^2
Simplifying the equation:
0 = 5.00v0 - 122.5
Rearranging the equation to solve for v0:
v0 = 122.5 / 5.00 ≈ 24.5 m/s
So, the initial speed of the egg is approximately 24.5 m/s.
Part B: How high does it rise above its starting point?
To find the height the egg rises above its starting point, we can use the equation for vertical displacement:
Δy = v0 * t - (1/2) * g * t^2
Using the values we know: v0 ≈ 24.5 m/s and t = 5.00 s, we can solve for Δy:
Δy = 24.5 * 5.00 - (1/2) * (-9.8) * (5.00)^2
Δy = 122.5 + 122.5
Δy = 245.0 m
Therefore, the egg rises approximately 245.0 m above its starting point.
Part C: What is the magnitude of its velocity at the highest point?
At the highest point, the velocity becomes zero momentarily as it changes direction from upward to downward. The magnitude of velocity (speed) is equal to the initial speed, so the answer is approximately 24.5 m/s.
Part D: What is the magnitude of its acceleration at the highest point?
The acceleration at the highest point is the acceleration due to gravity, which remains constant throughout the motion. Its magnitude is approximately 9.8 m/s^2.
Part E: What is the direction of its acceleration at the highest point?
The direction of acceleration at the highest point is always downward. It is consistent with the downward direction of the acceleration due to gravity, acting as an acceleration force on the egg. Hence, the direction of acceleration is down.