Mark ItIn a circus, a performer throws an apple towards a hoop held at 45 m45 m height by another performer standing on a high platform (see figure below). The thrower aim s for the hoop and throws the apple with a speed of 24 m s−124 m s-1. At the exact moment that the thrower released the apple, the other performer drops the hoop. The hoop falls straight down. At what height above the ground does the apple go through the hoop?
Question
Mark ItIn a circus, a performer throws an apple towards a hoop held at 45 m45 m height by another performer standing on a high platform (see figure below). The thrower aim s for the hoop and throws the apple with a speed of 24 m s−124 m s-1. At the exact moment that the thrower released the apple, the other performer drops the hoop. The hoop falls straight down. At what height above the ground does the apple go through the hoop?
Solution
To solve this problem, we need to understand that both the apple and the hoop are under the influence of gravity. This means they will both accelerate towards the ground at the same rate, which is approximately 9.8 m/s².
The hoop is dropped from a height of 45m, and the apple is thrown towards it. The question is asking at what height the apple will pass through the hoop.
Since they both fall at the same rate, the apple will meet the hoop at the same time the hoop has fallen to the height of the apple's trajectory.
We can use the equation of motion to find the time it takes for the hoop to hit the ground, which is the same time the apple will take to hit the hoop.
The equation is: h = 0.5gt², where h is the height (45m), g is the acceleration due to gravity (9.8 m/s²), and t is the time.
Rearranging for t gives: t = sqrt((2h)/g) = sqrt((2*45m)/9.8 m/s²) = 3.03 seconds.
Now we know the time, we can find the height at which the apple will meet the hoop. The apple is thrown upwards, so its height as a function of time is given by the equation: h = ut - 0.5gt², where u is the initial upward speed of the apple (24 m/s), g is the acceleration due to gravity (9.8 m/s²), and t is the time (3.03 seconds).
Substituting these values in gives: h = 24m/s * 3.03s - 0.59.8 m/s²(3.03s)² = 72.72m - 44.7m = 28.02m.
So, the apple will go through the hoop at a height of approximately 28.02m above the ground.
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