You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground, it is traveling at 25.0 m/s upward. Use the work–energy theorem to find (a) the rock’s speed just as it left the ground and (b) its maximum height.

(a) 30.32 m/s

(b) 46.88 m

Explanation:

Given:

Weight of the rock () = 3.00 N

Velocity at height 15.0 m (v) = 25 m/s

Displacement of the rock (S) = 15.0 m

(a)

Let the velocity at the start be 'u'.

The mass of the rock is given as:

Now, the work energy theorem states that the work done by the net force is equal to the change in the kinetic energy of the body.

Here, the net force is gravitational force acting on the rock vertically downward and the displacement of the rock is vertically upward. Therefore, the work done by gravity is negative and given as:

Now, change in kinetic energy is given as:

Plug in the given values and simplify. This gives,

From work-energy theorem:

Therefore, the speed with which it leaves the ground is 30.32 m/s.

(b)

Let the maximum height reached be 'h'.

We know that, at maximum height, the speed is 0 m/s.

Therefore, final speed (v) = 0 m/s

Initial speed (u) = 30.32 m/s

The work done by gravity is given as:

The change in kinetic energy is given as:

Again from work-energy theorem, we have:

Hence, the maximum height reached is 46.88 m.

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

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explanation: