Question

The baseball weighs 1 kg. The pitcher throws the ball at a velocity of 10 m/s straight up. Neglecting air resistance, what is the net force that acts on the baseball when it is HALF WAY to the top of the path

Answer 1

Answer:

9.81 N

Explanation:

Given that a baseball weighs 1 kg. The pitcher throws the ball at a velocity of 10 m/s straight up. Neglecting air resistance. The maximum height reached will be calculated by using the formula

V^2 = U^2 - 2gH

U = 10 m/s

g = 9.8m/s^2

At maximum height, V = 0

Substitute u and g into the formula

0 = 10^2 - 2 × 9.8 × H

19.6H = 100

H = 100/19.6

H = 5.1 m

The Kinetic energy on the ball will be

K.E = 1/2mv^2

K.E = 1/2 × 1 × 10^2

K.E = 1/2 × 100

K.E = 50 J

But energy = work done

WD = Force × distance (height)

The force that acts on the baseball when it is HALF WAY to the top of the path will be

F × 5.1/2 = 50

F = 100/5.1

F = 19.61 N

The weight acting downward will be

W = mg

W = 1 × 9.8

W = 9.8 N

The net force acting on the ball will be

Net force = F - W

Net force = 19.61 - 9.8

Net force = 9.81 N

Therefore, the net force that acts on the baseball when it is HALF WAY to the top of the path is 9.81 N