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 34.0 m below its starting point at a time 5.00 s after it leaves the thrower's hand. Air resistance may be ignored.
t = time to peak
1/2 g t^2 + 34 = 1/2 g (5 - t)^2
g t^2 + 68 = g (25 - 10t + t^2)
68 = g (25 - 10t)
thrown velocity = g t
To solve this problem, we can use the laws of motion and the kinematic equations. We are given the following information:
- The egg is thrown nearly vertically upward, meaning its initial velocity is positive.
- The egg just misses the cornice on the way down, which means its maximum height is less than the height of the cornice.
- At a certain point, the egg is 34.0 m below its starting point after 5.00 seconds.
To determine the initial velocity of the egg, we can use the equation for displacement:
Δy = v₀t + (1/2)at²
Where:
- Δy is the change in height (height above or below the starting point).
- v₀ is the initial velocity.
- t is the time.
- a is the acceleration (which is -9.8 m/s² due to gravity).
Since the egg is thrown vertically upward, the acceleration is in the opposite direction of the initial velocity. Therefore, we have a negative acceleration, making the equation:
-Δy = v₀t - (1/2)gt²
We are given that the egg is 34.0 m below its starting point after 5.00 seconds. Plugging in the values, the equation becomes:
-34.0 m = v₀(5.00 s) - (1/2)(9.8 m/s²)(5.00 s)²
Simplifying the equation:
-34.0 m = 5.00v₀ - 12.25 m
Rearranging the terms to solve for v₀:
5.00v₀ = -34.0 m + 12.25 m
5.00v₀ = -21.75 m
v₀ = (-21.75 m) / 5.00
v₀ ≈ -4.35 m/s
Therefore, the initial velocity of the egg is approximately -4.35 m/s (negative due to its upward direction).