Question

An elevator without a ceiling is ascending with a constant speed of 10 m/s. A boy on the elevator throws a ball directly upward, from a height of 2.0 m above the elevator floor, just as the elevator floor is 38 m above the ground. The initial speed of the ball with respect to the elevator is 24 m/s.

(a) What is the maximum height attained by the ball?

Answer 1

The question is about finding the maximum height a ball reaches after being thrown from a rising elevator. Using kinematic equations, we first find the max height above the floor of the elevator and then add the elevator's height above the ground to obtain the total maximum height attained by the ball.

Step-by-step explanation:

The student's question is about determining the maximum height attained by a ball thrown upward from an elevator ascending with constant velocity. The initial height of the ball above the ground is the sum of the elevator's height (38 m) and the height of the ball above the elevator floor (2 m). So, the initial height above ground is 38 m + 2 m = 40 m. Using the kinematic equations, we can calculate the maximum height the ball will reach from the elevator's floor.

The kinematics equation for vertical motion without air resistance is:

• vf^2 = vi^2 + 2a(s - si)

Where:

• vf is the final velocity (0 m/s at the maximum height),
• vi is the initial velocity (24 m/s as thrown by the boy),
• a is the acceleration due to gravity (-9.81 m/s^2 downwards),
• s is the final position (the maximum height),
• si is the initial position (2 m above the elevator floor).

Plugging in the values and solving for s gives us:

• 0 = (24 m/s)^2 + 2(-9.81 m/s^2)(s - 2 m)

From which we find the maximum height above the elevator floor. To find the height above the ground, we add the initial 40 m to the result.