Question

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.

Answer 1

The ball will reach a height of 50 feet.

Step-by-step explanation:

To determine how high the ball will go, we can use the kinematic equation for vertical motion:

Final velocity squared = Initial velocity squared + 2 * acceleration * displacement

First, let's find the final velocity when the ball reaches its maximum height. Since the ball is thrown vertically upwards, its final velocity at the highest point will be zero. Therefore, we can rewrite the equation as:

0 = (40 ft/s)^2 - 2 * (-32 ft/s^2) * displacement

Now, solve for the displacement:

displacement = (40 ft/s)^2 / (2 * 32 ft/s^2) = 50 ft

So, the ball will reach a height of 50 feet.

Answer 2

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.