In pulling two identical carry-on bags through the airport, Mr. Myers and his 13 year old grandson, Vincent, do the same amount

200*2N+500*5S=(700)V
<span>400N-2500N=700V </span>
<span>determine V. A negative N indicates S</span>

We apply the formula S=½gt², where S represents the vertical distance of 74.3m, and g is the acceleration due to gravity valued at 9.8 m/s². Therefore, t=√(2S/g)=3.884 seconds. Consequently, the car remains airborne for around 3.894 seconds. Its horizontal journey is vt=20×3.894=77.88m.

The ball was released from a height of 20 meters

Explanation:

The scenario is as follows:

1. A ball drops from the edge of a cliff.

2. Upon reaching the ground, the energy held in its gravitational potential energy transforms entirely into kinetic energy.

   This implies K.E = P.E.

3. The ball impacts the ground at a speed of 20 m/s.

4. The gravitational field strength noted is 10 N/kg.

<pOur goal is to ascertain the height from which the ball was dropped.

<pSince the ball was dropped from a cliff, its initial velocity is 0.<p→ K.E =

where v is the final velocity, is the initial velocity, and m is the mass.

<p→ v = 20 m/s and = 0 m/s.<p→ K.E =

→ K.E =

→ K.E = 200 m joules when the ball strikes the ground.

<p→ P.E = mg h

where g is the gravitational field strength, m is mass, and h signifies height.

<p→ g = 10 N/kg.<p→ P.E = m(10)(h)

→ P.E = 10m h joules.

<p→ P.E = K.E.

→ 10m h = 200 m.

Dividing through by 10m yields:

→ h = 20 meters.

The ball was released from a height of 20 meters.

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