Aball is thrown vertically upward with an initial speed of 40 m/s. how high is the ball above the ground when it stops

80m

Explanation:

According to the equation of motion, v²= u²+2as

Where v is the final velocity of the body at maximum height= 0m/s

u is the initial velocity of the body = 40m/s

a = -g (negative acceleration due to gravity since the object is thrown upward) = -10m/s²

S = H = the height of the ball above the ground

Substituting the values into the equation given, we will have;

0² = 40²-2(10)H

0 = 1600-20H

-1600 = -20H

H = -1600/-20

H = 80m

The height of the ball above the ground where it stops will be 80m

0.19 m/s and 0.45 m/s