A juggler performs in a room with a ceiling 2m abouve hand level. A. what is the maximum upward speed she can give the ball without hitting the celing.
B. what is the time spent in the air before the ball is caught?
A.
Equate potential and kinetic energies:
potential energy = mgh = mg(2) N-m
where g = acceleration due to gravity = 9.8 m/s²
kinetic energy = (1/2)mv²
Equate energies:
mgh = (1/2)mv²
solve for v (m/s)
B. t = v/g (deceleration by gravity)
(A) 6.3m/s
(B)1.3 s
To find the maximum upward speed of the ball, we can use the concept of projectile motion. We know that the displacement in the y-direction (vertical direction) will be the height of the ceiling, which is 2m.
A. To calculate the maximum upward speed, we can use the following equation of motion:
v_f^2 = v_i^2 + 2aΔy
where:
v_f is the final velocity (which will be 0 as the ball reaches maximum height),
v_i is the initial velocity (the speed given to the ball),
a is the acceleration due to gravity (-9.8 m/s^2), and
Δy is the displacement in the y-direction (which is 2m).
Substituting the known values into the equation, we have:
0 = v_i^2 + 2(-9.8)(2)
Simplifying the equation, we get:
0 = v_i^2 - 39.2
Now, solving for v_i (the maximum upward speed):
v_i^2 = 39.2
v_i = sqrt(39.2)
Using a calculator, we can determine that the maximum upward speed is approximately 6.26 m/s.
B. To find the time spent in the air before the ball is caught, we can use another equation of motion:
Δy = v_i*t + (1/2)*a*t^2
where:
Δy is the displacement in the y-direction (2m),
v_i is the initial velocity (the speed given to the ball),
a is the acceleration due to gravity (-9.8 m/s^2), and
t is the time spent in the air.
Substituting the known values into the equation, we have:
2 = v_i*t + (1/2)*(-9.8)*t^2
Simplifying the equation, we get:
0 = -4.9t^2 + v_i*t - 2
This is a quadratic equation, which can be solved by factoring, using the quadratic formula, or by using a graphing calculator.