Answer:
38.1 m/s
Step-by-step explanation:
To calculate the initial speed at which the ball must be thrown to reach a maximum height of 74 m, we can use the principles of projectile motion and the conservation of energy.
When the ball reaches its maximum height, all of its initial kinetic energy is converted into potential energy. We can equate these energies using the following equation:
Initial kinetic energy = Potential energy at maximum height
The initial kinetic energy is given by:
KE_initial = (1/2) * m * v^2
Where:
m is the mass of the ball, and
v is the initial velocity (speed) at which the ball is thrown.
The potential energy at maximum height is given by:
PE_max = m * g * h
Where:
g is the acceleration due to gravity (approximately 9.8 m/s^2), and
h is the maximum height reached by the ball (74 m).
Setting the two equations equal to each other:
(1/2) * m * v^2 = m * g * h
We can cancel out the mass (m) on both sides:
(1/2) * v^2 = g * h
Simplifying further:
v^2 = 2 * g * h
Finally, taking the square root of both sides to solve for v:
v = √(2 * g * h)
Substituting the given values:
v = √(2 * 9.8 m/s^2 * 74 m)
= √(1451.2 m^2/s^2)
≈ 38.1 m/s
Therefore, the ball must be thrown with a speed of approximately 38.1 m/s to reach a maximum height of 74 m.