with what minimum speed must a ball be thrown straight up in order to reach a height of 34 meters above the launch position?
v = Vi - 9.81 t
at top v = 0
so
t = Vi/9.81 at top
34 = 0 + Vi t - (9.81/2) t^2
34 = Vi^2/9.81 - Vi^2 /(2*9.81)
34 = Vi^2/(2*9.81)
Vi = 25.8 m/s
To determine the minimum speed required for a ball to reach a certain height when thrown straight up, we can use the principle of conservation of energy.
The initial energy of the ball is solely in the form of kinetic energy due to its motion. As it reaches the maximum height, this kinetic energy is converted into potential energy, given by the equation:
Potential Energy (PE) = m * g * h
Where:
m is the mass of the ball
g is the acceleration due to gravity (approximately 9.8 m/s^2 near the Earth's surface)
h is the height above the launch position
At the maximum height, the ball momentarily comes to rest before falling back down. Therefore, the potential energy at the maximum height is equal to the initial kinetic energy. So, we can set the potential energy equal to the initial kinetic energy:
Initial Kinetic Energy (KE) = PE
Since the ball is thrown straight up, its final velocity at the maximum height will be zero. The kinetic energy can be calculated using the formula:
Kinetic Energy (KE) = 1/2 * m * v^2
Where:
m is the mass of the ball
v is the initial velocity or speed of the ball
We can set the initial kinetic energy equal to the potential energy equation:
1/2 * m * v^2 = m * g * h
Simplifying the equation:
1/2 * v^2 = 2 * g * h
v^2 = 4 * g * h
v = √(4 * g * h)
Plugging in the values for g = 9.8 m/s^2 and h = 34 meters, we can calculate the minimum speed required for the ball:
v = √(4 * 9.8 * 34)
v ≈ √(1323.2)
v ≈ 36.36 m/s
Therefore, the minimum speed at which the ball needs to be thrown straight up is approximately 36.36 m/s to reach a height of 34 meters above the launch position.