Question

A 70 kg person is at the top of a 10 m tall slide. Ignoring friction, what is his or her speed at the halfway point where he or she is 5 m from the bottom of the slide? What is his or her speed at the bottom of the slide?

Answer 1

Answer on Question #48990, Physics, Mechanics - Kinematics - Dynamics

A 70 kg person is at the top of a 10 m tall slide. Ignoring friction, what is his or her speed at the halfway point where he or she is 5 m from the bottom of the slide? What is his or her speed at the bottom of the slide?

By the law of energy conservation:

$$E _ {p} = E _ {k}$$ $$m g h = \frac {m v ^ {2}}{2} \rightarrow v = \sqrt {2 g h}$$

At the halfway the speed is:

$$v _ {\frac {1}{2}} = \sqrt {2 \cdot 9 . 8 \frac {m}{s ^ {2}} \cdot 5 m} = 9. 9 \frac {m}{s}$$

At the bottom:

$$v _ {1} = \sqrt {2 \cdot 9 . 8 \frac {m}{s ^ {2}} \cdot 1 0 m} = 1 4 \frac {m}{s}$$

Answer: $v_{\frac{1}{2}} = 9.9 \frac{m}{s}$

$$v _ {1} = 1 4 \frac {m}{s}$$

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