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
