A football is thrown horizontally with an initial velocity of (16.6 m/s)x^. ignoring air resistance, the average acceleration of

Question

nadya68 [22]

3 years ago

12

A football is thrown horizontally with an initial velocity of (16.6 m/s)x^. ignoring air resistance, the average acceleration of

the football over any period of time is (-9.81 m/s2)y^.find the magnitude of the velocity at this time.

Physics

2 answers:

Andrei [34K] · Answer 1 · 2022-06-25T19:55:56+03:00

Solution:

According to the equations for 1-D kinematics. The only change to them is that instead one equation that describes general motion.

So we will have to use the equations twice: once for motion in the x direction and another time for the y direction.

v_f=v_o + at ……..(a)

[where v_f and v_o are final velocity and initial velocity, respectively]

Now ,

Initially, there was y velocity, however gravity began to act on the football, causing it to accelerate.

Applying this value in equation (a)

v_yf = at = -9.81 m/s^s * 1.75 = -17.165 m/s in the y direction

For calculating the magnitude of the equation we have to square root the given value

(16.6i - 17.165y)

\\ \left | V \right |=sqrt{16.6^{2}+17.165^{2}}\\ = \sqrt{275.56+294.637225}\\= \sqrt{570.197225}\\= 23.87[/tex]

rosijanka [135] · Answer 2 · 2022-06-25T21:42:09+03:00

The football's speed, at any period of time, is given by $v= \sqrt{(16.6)^2 + (9.81 \cdot t)^2}$ .

<h3>Further explanation</h3>

It seems some information is missing on the question, however we will try to answer the question the best as possible. Assuming that the football is thrown horizontally, let's say, from the edge of a cliff, then the football would be in what is called a "free fall". Free falling objects are those in which the only force which acts upon those bodies is gravity.

As the football falls, it gains velocity in the vertical direction (since gravity keeps pushing the object down), while its horizontal velocity remains constant (since we are ignoring air resistance, which is a very logical assumption). We can write this in equations as:

$v_x = 16.6$

$v_y = -9.81 \cdot t$

Where $v_x$ and $v_y$ are the horizontal and vertical velocities respectively. Speed is the magnitude of the velocity vector, and we can compute it in the following way:

$V(t)= \sqrt{(v_x)^2 + (v_y)^2} =\sqrt{(16.6)^2 + (9.81 \cdot t)^2}$

The above expression is useful since allows us to compute the speed of the object at any time, let's compare the velocity of the object at 1, 2, and 3 seconds:

$V(1)= \sqrt{(16.6)^2 + (9.81 \cdot 1)^2} = 19.28$ <em>m/s</em>