A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he releases it. Use energy to find The ball's speed as it passes the window on its way down and speed of impact on the ground.

Kinetic energy on the way up = kinetic energy on the way down, ignoring air-resistance.

Therefore speed going up = speed going down = 10 m/s.

Energy at window level = (1/2)mv0²+mgH
Energy at ground level = (1/2)mv²
Equate energies and solve for v.

Calculated it comes out to be 22

Kinematics can also be used

To find the ball's speed as it passes the window on its way down and the speed of impact on the ground, we can use the principle of conservation of energy.

Step 1: Find the potential energy of the ball at the window.
The potential energy (PE) of an object is given by the formula PE = m * g * h, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
PE = m * g * h
Since the mass of the ball is not given, we can assume it to be 1 kg (as it will cancel out in the final calculations).
PE = 1 kg * 9.8 m/s^2 * 20 m
PE = 196 joules

Step 2: Find the kinetic energy of the ball as it passes the window on its way down.
The kinetic energy (KE) of an object is given by the formula KE = 0.5 * m * v^2, where m is the mass and v is the velocity.
Since we assumed the mass of the ball to be 1 kg, the equation becomes:
KE = 0.5 * 1 kg * v^2
We'll denote the velocity as v1.

Step 3: Apply the conservation of energy principle.
According to the principle of conservation of energy, the total energy of the system remains constant. Therefore, the sum of the potential energy and kinetic energy at the highest point (when the ball was tossed) must be equal to the sum of the potential energy and kinetic energy at the window.
PE + KE = PE + KE1
196 joules + KE = 196 joules + KE1

Step 4: Substitute the potential and kinetic energy values into the equation.
196 joules + 0.5 * 1 kg * v1^2 = 196 joules + KE1
0.5 * v1^2 = KE1

Step 5: Find the speed of the ball as it passes the window on its way down.
To find the speed (v1), we'll set KE1 equal to the potential energy at the window and rearrange the equation:
0.5 * v1^2 = 196 joules
v1^2 = 196 joules * 2
v1^2 = 392 joules
v1 = √(392 joules)
v1 ≈ 19.8 m/s

Step 6: Find the speed of impact on the ground.
The speed of impact on the ground can be found using the principle of conservation of energy by equating the potential energy at the highest point (when the ball was tossed) to the kinetic energy at the impact.
PE = KE_impact
196 joules = 0.5 * m * v_impact^2
We'll denote the velocity at impact as v2.

Step 7: Determine the height from which the ball is dropped.
We need to find the maximum height reached by the ball before it strikes the ground. This height (h') can be calculated using the equation:
PE = m * g * h'
196 joules = 1 kg * 9.8 m/s^2 * h'
h' = 20 m

Step 8: Find the speed of impact on the ground.
To find the speed (v2), we'll set the potential energy at the highest point equal to the kinetic energy at impact and rearrange the equation:
0.5 * m * v_impact^2 = m * g * h'
0.5 * v2^2 = 9.8 m/s^2 * 20 m
0.5 * v2^2 = 196 m^2/s^2
v2^2 = 392 m^2/s^2
v2 = √(392 m^2/s^2)
v2 ≈ 19.8 m/s

Therefore, the ball's speed as it passes the window on its way down is approximately 19.8 m/s, and the speed of impact on the ground is also approximately 19.8 m/s.

To find the ball's speed as it passes the window on its way down and the speed of impact on the ground, we can use the principle of conservation of mechanical energy.

At the highest point of its trajectory, the ball will have gravitational potential energy but no kinetic energy. As the ball falls, gravitational potential energy is converted to kinetic energy.

First, let's determine the velocity of the ball when it reaches its highest point. We can use the equation for potential energy:

PE = mgh

Where PE is the potential energy, m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the ground.

Given that the ball is 20 m above the ground, the potential energy when released is:

PE = (mass of the ball) * (9.8 m/s^2) * 20 m

Next, let's determine the ball's velocity when reaching the highest point. At the highest point, all the potential energy is converted to kinetic energy. The equation for kinetic energy is:

KE = (1/2) * m * v^2

Where KE is the kinetic energy and v is the velocity of the ball.

Setting the potential energy equal to the kinetic energy:

(mass of the ball) * (9.8 m/s^2) * 20 m = (1/2) * (mass of the ball) * v^2

Simplifying the equation:

(9.8 m/s^2) * 20 m = (1/2) * v^2

The mass of the ball cancels out:

v^2 = (9.8 m/s^2) * 20 m * 2

Now we solve for v:

v^2 ≈ 392 m^2/s^2

Taking the square root of both sides:

v ≈ √(392 m^2/s^2)

v ≈ 19.80 m/s

Therefore, the velocity of the ball as it passes the window on its way down is approximately 19.80 m/s.

To find the speed of impact on the ground, we can use the fact that the change in potential energy is equal to the change in kinetic energy. Initially, the ball has only potential energy, and at impact, it has only kinetic energy.

Let's calculate the potential energy when the ball is 20 m above the ground:

PE = (mass of the ball) * (9.8 m/s^2) * 20 m

Next, let's calculate the kinetic energy at impact. The velocity at impact will be the same as the velocity when the ball passed the window on its way down:

KE = (1/2) * (mass of the ball) * v^2

Plugging in the values we know:

KE = (1/2) * (mass of the ball) * (19.80 m/s)^2

Since the total energy is conserved, we have:

PE = KE

(mass of the ball) * (9.8 m/s^2) * 20 m = (1/2) * (mass of the ball) * (19.80 m/s)^2

Simplifying the equation:

(9.8 m/s^2) * 20 m = (1/2) * (19.80 m/s)^2

Now we solve for the speed of impact:

Speed of impact ≈ √(2 * (9.8 m/s^2) * 20 m)

Speed of impact ≈ √(392 m^2/s^2)

Speed of impact ≈ 19.80 m/s

Therefore, the speed of impact on the ground is approximately 19.80 m/s.