A ball is thrown vertically upwards from a height of 6 feet with an initial velocity of 40 feet per second. How high will the ball go? Note that the acceleration of the ball is given by feet per (second) 2. Round your answer to the four decimal places.

The maximum height that the ball reaches is sf = 31 feet from the ground

Step-by-step explanation:

Solution:-

- A ball is thrown vertically upwards from an initial elevation of si = 6 feet from the ground.

- The velocity with which the ball was thrown up, vi = 40 ft/s

- We can determine the maximum height of the ball when it is thrown vertically up by using the 3rd kinematic equation of motion.

vf^2 - vi^2 = 2*g * (sf - si)

Where,

vf : The final velocity of the ball, for maximum height it is = 0

sf : The final height of the ball from the ground

g : Gravitational constant = - 32 ft/s^2

0 - 40^2 = - 2*32 * (sf - 6)

sf - 6 = 25

sf = 31 feet

- The maximum height that the ball reaches is sf = 31 feet from the ground.