Question

A batter hits a baseball m = 0.49 kg from rest with a force of f = 88 n, the ball is in contact with the bat for t = 0.11 s. it is observed that the ball leaves the bat with an angle of θ = 24 degrees with respect to the horizontal. how long was the ball in the air if it was hit from the top of a cliff with a height of h = 142 m?

Answer 1

The ball was in the air for approximately 8.86 seconds.

Step-by-step explanation:

To solve this problem, we can use the principle of conservation of mechanical energy. The initial mechanical energy of the ball is equal to its potential energy when it was at the top of the cliff, which can be calculated using the formula PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the cliff. The final mechanical energy of the ball is equal to its kinetic energy when it was in the air, which can be calculated using the formula KE = 1/2 * mv^2, where v is the velocity of the ball when it leaves the bat. Since mechanical energy is conserved, we can equate the initial mechanical energy to the final mechanical energy:

mgh = 1/2 * mv^2

From the given information, we have m = 0.49 kg, g = 9.8 m/s^2, h = 142 m, and v = 88 m/s. Plugging these values into the equation, we can solve for t, the time the ball was in the air:

t = 2 * sqrt((h * m) / (g * v^2))

Substituting the given values, we get t ≈ 8.86 seconds. Therefore, the ball was in the air for approximately 8.86 seconds.