A ball is launched at an upward angle of 45 degrees from the top of a cliff 165 meters high, with a speed of 180 m/s. Use conservation of energy to find how fast it will be traveling when it hits the ground.
final ke= initial KE + initialGPE
189 m/s
To find the speed at which the ball hits the ground, we can use the principle of conservation of energy. At the top of the cliff, the ball has potential energy due to its height, and as it falls, this potential energy is converted into kinetic energy.
The formula for gravitational potential energy is given by:
PE = mgh
where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.
In this case, we are given the height of the cliff (h = 165 meters) but not the mass of the ball. However, we don't need the mass of the ball to solve the problem because it cancels out in our calculations.
The kinetic energy (KE) is given by the formula:
KE = 0.5 * m * v²
where KE is the kinetic energy and v is the velocity.
According to the principle of conservation of energy, the potential energy at the top of the cliff should be equal to the kinetic energy just before reaching the ground.
Therefore, we can equate the potential energy to the kinetic energy:
mgh = 0.5 * m * v²
Now, we can cancel out the mass (m) on both sides of the equation:
gh = 0.5 * v²
Plugging in the values we know, we have:
(9.8 m/s²) * (165 m) = 0.5 * v²
Now we can solve for v:
1597 = 0.5 * v²
Dividing both sides by 0.5:
v² = 3194
Taking the square root of both sides:
v ≈ 56.5 m/s
Therefore, the ball will be traveling at approximately 56.5 m/s when it hits the ground.