A 35 kg child starting from rest slides down a hill with a vertical height of 20m. What is the child's speed at the bottom of the hill?

How do you solve this???

PE->KE

mgh=mv²/2

v=sqrt(2gh)

To solve this problem, you can make use of the principles of conservation of energy and the work-energy theorem.

Step 1: Calculate the gravitational potential energy (GPE) of the child at the top of the hill using the formula:
GPE = m * g * h
where m is the mass of the child (35 kg), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the vertical height of the hill (20 m).

GPE = 35 kg * 9.8 m/s² * 20 m
GPE = 6860 Joules

Step 2: At the bottom of the hill, the child's gravitational potential energy is fully converted into kinetic energy. Therefore, the kinetic energy (KE) of the child at the bottom of the hill is equal to the GPE calculated in step 1:
KE = GPE
KE = 6860 Joules

Step 3: Now, using the formula for kinetic energy:
KE = 0.5 * m * v²
where v is the velocity or speed of the child at the bottom of the hill.

Rearranging the formula, we get:
v² = (2 * KE) / m
v² = (2 * 6860 Joules) / 35 kg
v² = 392 Joules / kg

Step 4: Taking the square root of both sides of the equation, we find:
v = √(392 Joules / kg)

Calculating the square root, we get:
v ≈ 6.26 m/s

Therefore, the child's speed at the bottom of the hill is approximately 6.26 m/s.