Question

Nancy is playing on a ladder and slide. The entry point of the slide is 3.0 meters above the ground and the slide is inclined at an angle of 30.0° with the horizontal. What is Nancy’s displacement each time she slides down from the top?

Answer 1

Answer on Question 59425, Physics, Mechanics, Relativity

Question:

Nancy is playing on a ladder and slide. The entry point of the slide is $3.0 \, m$ above the ground and the slide is inclined at an angle of $30.0{}^{\circ}$ with the horizontal. What is Nancy's displacement each time she slides down from the top?

Solution:

Here's the sketch of our task:

The displacement is equal to the length of the slide $l$ that Nancy travels when she slides down from the top. Then, we can find the displacement from the right triangle:

$$\sin \theta = \frac {h}{l},$$

here, $\theta$ is the angle of inclination of the slide, $h$ is the height of the slide, $l$ is the length of the slide.

From the last formula we can calculate the Nancy's displacement:

$$\text {Displacement} = l = \frac {h}{\sin \theta} = \frac {3 . 0 m}{\sin 3 0 . 0 {}^ {\circ}} = \frac {3 . 0 m}{0 . 5} = 6. 0 m.$$

Answer:

Displacement $= 6.0m$

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