A juggler performs in a room with a ceiling 2.71 m above hand level. What is the maximum upward speed (in m/s) which she can give a ball without letting it hit the ceiling?
_________m/s
To find the maximum upward speed the juggler can give a ball without letting it hit the ceiling, we need to consider the motion of the ball and the constraints provided.
Since the ball will experience free fall, we can use the equations of motion to solve for the maximum upward speed. In this case, we can use the kinematic equation:
v_f^2 = v_i^2 + 2aΔy
Where:
v_f is the final velocity (which is zero at the highest point)
v_i is the initial velocity (the maximum upward speed)
a is the acceleration due to gravity (approximately 9.8 m/s^2)
Δy is the distance between the initial position and the highest point (the height of the ceiling)
By rearranging the equation, we can solve for the maximum upward speed (v_i):
v_i = sqrt(-2aΔy)
Substituting the given values, we have:
v_i = sqrt(-2 * 9.8 m/s^2 * 2.71 m)
Calculating this expression, we find:
v_i ≈ 6.58 m/s
Therefore, the maximum upward speed the juggler can give a ball without letting it hit the ceiling is approximately 6.58 m/s.