circus juggler performs an act with balls that he tosses with his right hand and catches with his left had. Each ball is launched at an angle of 75° and reaches a maximum height 88 cm above the launching height. If it takes the juggler 0.2 s to catch a ball with his left hand, pass to his right hand and toss it back into the air, what is the marimum number of balls he can juggle?
Honors Physics Fall Exam 2014
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Honors Physics Fall Exam 2014
I think you are really asking how long does it take to go up 0.88 meters and how long to come back down. This is presumably on earth where the acceleration due to gravity is -9.81 m/s^2
So what was initial velocity component upward?
(1/2)m Vi^2 = m g h conservation of energy
Vi = sqrt (2 g h)
= sqrt (2 *9.81*.88)
= 4.16 meters/second
now the AVERAGE speed upward is half of that so
Vav = 2.08 m/s
so the time to rise .88 meters is
trise = 0.88/2.08 = 0.424 seconds
fall time is the same so the total time in the air for this toss is
0.847 second
To determine the maximum number of balls the juggler can juggle, we need to consider the time it takes for a ball to complete one full cycle: from being tossed with the right hand, reaching the maximum height, falling back down, being caught with the left hand, passed to the right hand, and then tossed again.
From the given information, we know that the time it takes for the juggler to catch, pass, and toss the ball is 0.2 seconds. This means that a complete cycle for a ball takes 0.2 seconds.
Now let's calculate the total time it takes for a ball to complete one cycle. The time for the ball to reach the maximum height is half of the total time taken for the full cycle. Therefore, the time for the ball to reach the maximum height is 0.2 seconds divided by 2, which is 0.1 seconds.
Next, we can use the equation of motion to calculate the time it takes for the ball to reach the maximum height:
h = (v^2 * sin^2(θ)) / (2 * g)
Where:
h = maximum height (88 cm = 0.88 m)
v = initial velocity
θ = launch angle (75°)
g = acceleration due to gravity (9.8 m/s^2)
Rearranging the equation to solve for v, we get:
v = √((2 * g * h) / sin^2(θ))
Substituting the given values, we can determine the initial velocity of the ball:
v = √((2 * 9.8 * 0.88) / sin^2(75°))
Calculating this value, we find that v ≈ 4.234 m/s.
Since the time taken for the ball to reach the maximum height is 0.1 seconds and the initial velocity is 4.234 m/s, we can calculate the initial vertical velocity (v_y) using the formula:
v_y = g * t
Where:
v_y = initial vertical velocity
g = acceleration due to gravity (9.8 m/s^2)
t = time taken for the ball to reach the maximum height (0.1 seconds)
Plugging in the values, we have:
v_y = 9.8 * 0.1
Calculating this, we find that v_y = 0.98 m/s.
Now, we know that to catch, pass, and toss the ball, the juggler needs a total time of 0.2 seconds. During this time, the ball will complete one full cycle. Therefore, the time for the ball to fall back down from its maximum height to the catching hand is the remaining time after reaching the maximum height:
t_fall = 0.2 seconds - 0.1 seconds
So, t_fall = 0.1 seconds.
Using this time and the initial vertical velocity (v_y), we can determine the distance the ball falls during this time using the formula:
d = v_y * t_fall + 0.5 * g * t_fall^2
Plugging in the values, we have:
d = 0.98 * 0.1 + 0.5 * 9.8 * (0.1)^2
Calculating this, we find that d ≈ 0.098 m.
Since the ball fell by a distance of approximately 0.098 m, the juggler needs to throw the ball up by this same distance to catch it again. Therefore, the total vertical distance the ball travels during one complete cycle is approximately 88 cm + 98 cm = 186 cm = 1.86 m.
Now, let's determine the total time it takes for the juggler to complete one full cycle (from the launch to the catch of the same ball). Since the ball is thrown up and caught in the air, we have two vertical distances to consider: the distance it takes to reach the maximum height (88 cm) and the distance it takes to fall back down (98 cm). Let's call this total time T.
T = t_up + t_down
To calculate t_up (the time it takes for the ball to reach the maximum height), we can use the equation of motion:
s = v * t + 0.5 * a * t^2
Where:
s = distance (88 cm = 0.88 m)
v = initial velocity (4.234 m/s)
a = acceleration due to gravity (-9.8 m/s^2, negative since it acts in the opposite direction to the initial velocity)
t = time (unknown)
Rearranging the equation to solve for t, we get a quadratic equation:
-0.5 * 9.8 * t^2 + 4.234 * t - 0.88 = 0
Solving this equation using the quadratic formula, we find two possible values for t: t = 0.084 s or t = 0.112 s.
Since we are interested in the total time for one full cycle, we want to consider the longer time, which is t = 0.112 s for reaching the maximum height (t_up).
Now, to calculate t_down (the time it takes for the ball to fall back down), we already know it is 0.1 seconds (from the original information given).
Therefore, the total time for one full cycle (T), considering the time it takes for the ball to reach the maximum height and fall back down, is:
T = t_up + t_down
T = 0.112 s + 0.1 s
T = 0.212 s
Since we know it takes 0.2 seconds for the juggler to catch, pass, and toss the ball, the juggler can complete approximately one full cycle in this time (0.212 s is slightly greater than 0.2 s).
Hence, the maximum number of balls the juggler can juggle is: 1.