Answer :
a) the child's speed halfway down the slide's vertical distance is 9.89 m/s
b)the child's speed three-fourths of the way down is 12.12 m/s
the problem we are dealing with is related to gravitational potential energy is the potential energy an enormous object has in connection to another gigantic object due to gravity. It is the potential energy related to the gravitational field, which is discharged when the objects drop toward each other. Gravitational potential energy increments when two objects are brought advanced separated.
Since we are given the problem of a 30.0 kg child starting from rest slides down a water slide with a vertical height of 10.0m.
Since we know that ΔPE=KE
=> mgΔh=mv²/2
as we know v=sqrt(2gΔh)
so halfway down the slide, Δh is 5 m, and gravity = 9.8 m/s²
Now substituting the values of Δh in the formula of the v
v=sqrt(2gΔh)
v= √2* 9.8 * 5
= √98
= 9.89 m/s
for the three fourth of the way down, Δh=7.5 m.
v= √2* 9.8 * 7.5
= 12.12 m/s
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