Answer:
1.8 m
Step-by-step explanation:
The motion from the top of the slide is virtually under gravity. The motion starts from rest. We use the equation of motion
[tex]v_f^2 = v_i^2 + 2as[/tex]
[tex]v_f[/tex] and [tex]v_i[/tex] are the final and initial velocities, respectively, a is the acceleration and s is the displacement.
For this motion, initial velocity is 0 m/s, final velocity is 6 m/s, a is the acceleration of gravity (9.8 m/s2) and s is the height, h.
[tex]v_f^2 = v_i^2 + 2gh[/tex]
[tex]6^2 = 0^2 + 2\times9.8h[/tex]
[tex]h=\dfrac{36}{19.6}=1.8 \text{ m}[/tex]
The maximum height that the slide can be is 1.8m.
Since the speed a child reaches at the bottom does not exceed 6.0 m/s.
So here the maximum height should be
[tex]6^2 = 0^2 + 2\times 9.8\\\\h = 3.6\div 19.6[/tex]
= 1.8m
Hence, The maximum height that the slide can be is 1.8m.
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