Suddenly losing his thirst, Mathman decides to check out the festivities, but just then, Frankenstein ran by him screaming, "Cheater, cheater, apple eater." It seems that Dracula had illegally been using his bicuspids to win at apple-bobbing. Frankenstein was furious and ran outside with the apple. He then heaved it skyward at a rate of 192 feet per second. If he released the apple at a height of 10 feet, how high did the apple go? When did it return to the ground?
Thanks
h = 10 + 192t - 16t^2
This reaches its peak at t = -192/-32 = 6
h(6) = 586 feet
h = 0 when t = 12.05 sec
To determine how high the apple went and when it returned to the ground, we can use the principles of projectile motion.
First, let's calculate how long it takes for the apple to reach its maximum height. We'll use the following equation:
t = (Vf - Vi) / g
Where:
t is the time of flight,
Vf is the final vertical velocity (0 ft/s when the apple reaches its maximum height),
Vi is the initial vertical velocity (192 ft/s),
g is the acceleration due to gravity (-32.2 ft/s^2).
Plugging in the values, we get:
t = (0 - 192) / -32.2
t ≈ 5.96 seconds
Since the apple reaches its maximum height halfway through its total flight time, it takes approximately 2.98 seconds to reach the maximum height.
Next, let's calculate the maximum height (hmax) the apple reaches. We'll use the following equation:
hmax = Vi * t + (1/2) * g * t^2
Plugging in the values, we get:
hmax = 192 * 2.98 + (1/2) * -32.2 * (2.98)^2
hmax ≈ 283.48 feet
Therefore, the apple reaches a maximum height of approximately 283.48 feet.
To calculate when the apple returns to the ground, we can use the equation for the total time of flight (T):
T = 2t
Plugging in the value of t, we get:
T = 2 * 5.96
T ≈ 11.92 seconds
Therefore, the apple returns to the ground after approximately 11.92 seconds.
In summary, the apple reaches a maximum height of approximately 283.48 feet and returns to the ground after approximately 11.92 seconds.