Physics, 25.09.2019 20:20 tangia

Asoccer ball is kicked straight upwards with an initial vertical speed of 8.0 m/s. we can ignore air resistance. how long does it take the ball to have a downwards speed of 4.0 m/s?

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Answer from: caleighpeters

The time the ball it takes to have the downward speed 4 m/s is 1.22 sec.

Explanation:

Lets explain the motion of the ball

The ball is kicked straight upwards with an initial vertical speed of 8.0 m/s

Its velocity decreases util it reaches the maximum height

At the maximum height its velocity is equal to zero

Then it fall down to the ground

We can ignore air resistance

Lets find the time it takes the ball to reach the maximum height

The acceleration of gravity is -9.8 m/s²

The initial velocity is 8 m/s upward

The final velocity is zero

So lets use the rule v = u + gt, where v is the final velocity, u is the initial

velocity, g is the acceleration of gravity and t is the time

u = 8 , v = 0 , g = -9.8

Substitute these value in the equation above

0 = 8 - 9.8 t

Add 9.8 t for both sides

9.8 t = 8

Divide both sides by 9.8

t = 40/49 seconds

The ball falls withe initial velocity zero and increases to reach 4 m/s

The velocity 4 m/s is downward

The acceleration of gravity is 9.8 m/s²

By using the same rule

u = 0 , v = 4 , g = 9.8

4 = 0 + 9.8 t

4 = 9.8 t

Divide both sides by 9.8

t = 20/49

The time the ball it takes to have the downward speed 4 m/s is the sum

of the two times above

The time the ball it takes = $\frac{40}{49}+\frac{20}{49}=\frac{60}{49}=1.22$ seconds

The time the ball it takes to have the downward speed 4 m/s is 1.22 sec.

Answer from: Quest

talk back to him about the sport and show him your expertise or leave the room

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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Asoccer ball is kicked straight upwards with an initial vertical speed of 8.0 m/s. we can ignore air...

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Physics, 07.02.2021 05:40

Can someone help me grasp the concept of solving this please