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

80m, assuming g=10m/s^2

Explanation:

40m/s will be reduced to 0m/s in 4 seconds. 4 seconds x 40m/s would be 160m up, but you will only get half of that because you decelerate linearly to 0m/s. This leaves you with 4 x 20 = 80m.

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