I utilized delta y= -1/2 gt^2 and calculated .60 seconds for time which I then p

Question

I utilized

delta y= -1/2 gt^2

and calculated .60 seconds for time which I then plugged into

delta x = v(initial) cos(theta) (time)

and calculated v(initial) = 4.35 m/s

then I utilized that for v(final) in

mg(h +1.8) = 1/2 v(final)^2 + g(1.8)

I removed m since it is in all parts of the equation

I then calculated 2.765433m and webassign.com says my answer is incorrect....

I also subtracted 1.8m to see if that would be the height they are looking for = .965m

But webassign also says that answer is incorrect

Explanation / Answer

1.8 = 4.9t^2
t = 0.6 sec.
to travel 2.61 m in 0.6 seconds the child came of the slide with a speed of
2.61/0.6 = 4.35 m/s

v^2 = 2gh
18.93= 19.6h
h = 0.965 meters

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.