Solving the Puzzle: Ball Thrown Upward

[iluvpandas]

Homework Statement

A ball is thrown straight up with an initial speed of 40 m/s. (Neglect air resistance.)
How high does it go? how long does it stay i the air

Homework Equations

i know the second part is 8 because its basically saying 40m every second divide by acc.(gravity) which is basically 10 and multiply by 2 one second going up and another going down. but I am completely lost on the first part

The Attempt at a Solution

im not sure but i think i use the formula for displacement? but when i tried the answer was wrong.

 

[rootX]

Write the three motion equations here. And then see what you are looking for.
What is the final velocity?
What is initial velocity?
What is distance? - not known
What is time? - not given
What is acceleration?

You should try equation with time in it when finding the distance.

 

[tomcue]

I would approach this problem by first identifying the known and unknown quantities. In this case, the known quantities are the initial speed (40 m/s) and the acceleration due to gravity (10 m/s^2). The unknown quantities are the height reached by the ball and the time it stays in the air.

To solve for the height, we can use the equation for displacement: s = ut + 1/2at^2, where s is the displacement (height), u is the initial speed, a is the acceleration, and t is the time.

Plugging in the known values, we get:
s = (40 m/s)(t) + 1/2(10 m/s^2)(t^2)
s = 40t + 5t^2

To solve for the time, we can use the equation for time: t = v/u, where v is the final velocity and u is the initial speed.

Since the ball reaches its maximum height at the top of its trajectory, the final velocity is 0 m/s. Thus, we get:
t = 0/40 m/s
t = 0 seconds

This means that the ball reaches its maximum height in 0 seconds, before falling back down. Therefore, the height reached by the ball is 0 meters.

To solve for the time it stays in the air, we can use the equation for time again:
t = v/u

Since the ball reaches its starting point again when it falls back down, the final velocity is equal to the initial velocity (40 m/s). Thus, we get:
t = 40 m/s / 40 m/s
t = 1 second

Therefore, the ball stays in the air for 1 second before falling back down.

In summary, the ball reaches a height of 0 meters and stays in the air for 1 second when thrown straight up with an initial speed of 40 m/s, neglecting air resistance.