Question

An 50 kg student runs and jumps onto a 25 kg tire swing, which is initially at rest, and rides the swing up into the air. The swing (and the student) then rises a vertical distance of 1.5 m. What was the initial speed, in m/s of the swing? Remember the acceleration due to gravity near Earth's surface, g = 9.81 m/s^2

A. 29 m/s

B. 5.4 m/s

C. 1.7 m/s

D. 15. m/s

E. 3.8 m/s


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Answer 1

Answer:

Option B, 5.4m/s

Explanation:

Using the formula v² = u ² - 2gh

v = 0(final velocity is zero at maximum height), u = ?, g = 9.81m/s², h = 1.5m

0² = u² - 2×9.81×1.5

0 = u² - 29.43

29.43 = u²

u = √29.43

= 5.42m/s