A softball pitcher swings a ball of mass 0.244 kg around a vertical circular path of radius 59.3 cm before releasing it from her hand. The pitcher maintains a component of force on the ball of constant magnitude 30.3 N in the direction of motion around the complete path. The speed of the ball at the top of the circle is 14.7 m/s. If the ball is released at the bottom of the circle, what is its speed upon release

its speed upon release is 26.05 m/s

Explanation:

Given that;

mass m = 0.244 kg

force F = 30.3 N

V1 = 14.7 m/s

r = 59.3 cm = 0.593 m

Vf = ?

we know that;

1/2mV1² + FπR = 1/2mVf²

so we substitute

[1/2×0.244×(14.7)²] + [30.3×π×0.593 = 1/2×0.244×Vf²

26.3629 + 56.4478 = 0.122Vf²

82.8107 = 0.122Vf²

Vf² = 82.8107 / 0.122Vf

Vf² = 678.7762

Vf = √678.7762

Vf = 26.05 m/s

Therefore its speed upon release is 26.05 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s