A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the speed of the ball just before it hits the catcher's glove 1.0 m above the ground? (g = 9.8 m/s2 and ignore air resistance)

V = sqrt(2gH)

The mass is irrelevant.
H is the height change, which is 100 m in this case.

To find the speed of the ball just before it hits the catcher's glove, we can use the principle of conservation of energy. The total mechanical energy (E) of the ball at the top of the drop (potential energy) is equal to the total mechanical energy of the ball just before it hits the catcher's glove (kinetic energy).

The potential energy (PE) of the ball at the top of the drop is given by the equation PE = mgh, where m is the mass of the ball (0.15 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the drop (101 m).

PE = 0.15 * 9.8 * 101
PE = 149.43 J

Since the total mechanical energy is conserved, the kinetic energy (KE) of the ball just before it hits the catcher's glove is also 149.43 J.

The kinetic energy (KE) of the ball is given by the equation KE = 1/2 mv^2, where m is the mass of the ball (0.15 kg), and v is the velocity of the ball just before it hits the catcher's glove.

149.43 = 1/2 * 0.15 * v^2
298.86 = 0.15 * v^2
v^2 = 298.86 / 0.15
v^2 = 1992.4
v = √1992.4
v ≈ 44.61 m/s

Therefore, the speed of the ball just before it hits the catcher's glove is approximately 44.61 m/s.

To find the speed of the ball just before it hits the catcher's glove, we can use the principle of conservation of energy.

Step 1: Find the potential energy of the ball at a height of 101 m.
The potential energy (PE) of an object is given by the equation PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.

Plugging in the values, we get:
PE = (0.15 kg)(9.8 m/s^2)(101 m)

Step 2: Find the kinetic energy of the ball just before it hits the catcher's glove.
The kinetic energy (KE) of an object is given by the equation KE = 1/2mv^2, where m is the mass of the object and v is the velocity/speed.

Since there is no change in the height of the ball (1.0 m), the potential energy at that height is negligible. Therefore, all the potential energy is converted into kinetic energy. Thus, we can equate the potential energy to the kinetic energy:

PE = KE

(0.15 kg)(9.8 m/s^2)(101 m) = 1/2(0.15 kg)v^2

Step 3: Solve for the velocity (v).
Rearrange the equation to solve for v:

v^2 = (2(0.15 kg)(9.8 m/s^2)(101 m)) / 0.15 kg

v^2 = 2(9.8 m/s^2)(101 m)

v^2 = 1989.6 m^2/s^2

v = √(1989.6 m^2/s^2)

v ≈ 44.6 m/s

Therefore, the speed of the ball just before it hits the catcher's glove is approximately 44.6 m/s.

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