A 33.5 kg child starting from rest slides down?

Question

A 33.5 kg child starting from rest slides down a water slide with a vertical height of 19.0 m. (Neglect friction.)
(a) What is the child's speed halfway down the slide's vertical distance?
m/s
(b) What is the child's speed three-quarters of the way down?
m/s

Answer 1

v = sqrt (2gd)
for a d will be 19*.5=9.5
a. sqrt (2 *9.8* 9.5)=13.65 m/s

for b d will be 19*.75=14.25
b. sqrt (2 *9.8* 14.25)=16.71m/s

Answer 2

This is the biology board...try the physics board.

Well....ok

Since friction is negligible, the child just as well be 'free-falling'.
Also, his acceleration will be uniform. Good ole simple kinematics.
a)
Half way down, the child has traveled 9.5 meters.
the acceleration is 9.8 m/s2

To determine the speed of the child at those spots, we need to know how long it took the child to get there.
X(orY)final = X(orY)initial + (Acceleration)(time)
9.5 m = 0 + 10t
9.5/9.8 = t = 0.97s (very close to 1)

So...we know he was travelling for one second when he hit the first checkpoint. Now we can calculate his speed.
Vfinal = Vinitial + 1/2 at^2
= 0 + 1/2 at^2
= 1/2 at^2
(1/2)(9.8)(1)^2
(0.5)(9.8)(1) = 4.9 meters per second.

Repeat this process for the next check point. Find the time it took to get there, then you can find the speed he is going. Good Luck.

Answer 3

CONSERVATION OF capability: Gravitational skill capability lost = mgd = Kinetic capability won= a million/2 mv^2 sparkling up for the fee: v = sqrt (2gd) observe that the mass cancels, so that's beside the point. observe additionally that by making use of making use of capability particularly than monitoring the forces, you do not might desire to difficulty your self with the path--purely the preliminary and intensely final positions. it must be vertical, diagonal, looped, or maybe if--the path is beside the point. for a), use 0.5 the given top for the drop, d. for b), use 3/4 the given top for d. you comprehend g. Plugnchug.

Answer 4

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