Physics, 16.10.2020 07:01 EMQPWE

Make a motion chart for a cannonball launched with an initial speed of 20m/s. Neglect drag and the initial height of the cannonball. After the ball lands on the ground enter zero for all speeds and heights if necessary. Use regular metric units (ie. Meters).

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Answer from: hinacat87

The speed and height at 0 sec are 20 m/s and 0 m.

The speed and height at 1 sec are 10.2 m/s and 15.1 m.

The speed and height at 2 sec are 0.4 m/s and 20.4 m.

The speed and height at 3 sec are 9.4 m/s and 15.9 m.

The speed and height at 4 sec are 19.2 m/s and 1.6 m.

Explanation:

Given that,

Initial speed = 20 m/s

Suppose, given time in charge are t= 0, 1, 2, 3, 4 sec

At t=0 sec,

We need to calculate the velocity and height

Using equation of motion

v=u-gt

Put the value into the formula

$v=20-9.8\times0$

$v=20\ m/s$

Now, using second equation of motion

$h=ut-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=20\times 0-\dfrac{1}{2}\times9.8\times0$

$h=0\ m$

At t=1 sec,

We need to calculate the velocity and height

Using equation of motion

v=u-gt

Put the value into the formula

$v=20-9.8\times1$

$v=10.2\ m/s$

Now, using second equation of motion

$h=ut-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=20\times1-\dfrac{1}{2}\times9.8\times1^2$

$h=15.1\ m$

At t=2 sec,

We need to calculate the velocity and height

Using equation of motion

v=u-gt

Put the value into the formula

$v=20-9.8\times2$

$v=0.4\ m/s$

Now, using second equation of motion

$h=ut-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=20\times2-\dfrac{1}{2}\times9.8\times2^2$

$h=20.4\ m$

At t=3 sec,

We need to calculate the velocity and height

Using equation of motion

v=u-gt

Put the value into the formula

$v=20-9.8\times3$

$v=-9.4\ m/s$

Negative sign shows the opposite direction of motion

Now, using second equation of motion

$h=ut-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=20\times3-\dfrac{1}{2}\times9.8\times3^2$

$h=15.9\ m$

At t=4 sec

We need to calculate the velocity and height

Using equation of motion

v=u-gt

Put the value into th formula

$v=20-9.8\times4$

$v=-19.2\ m/s$

Negative sign shows the opposite direction of motion

Now, using second equation of motion

$h=ut-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=20\times4-\dfrac{1}{2}\times9.8\times4^2$

$h=1.6\ m$

So, The ball has hit the ground at 5 sec.

Hence, The speed and height at 0 sec are 20 m/s and 0 m.

The speed and height at 1 sec are 10.2 m/s and 15.1 m.

The speed and height at 2 sec are 0.4 m/s and 20.4 m.

The speed and height at 3 sec are 9.4 m/s and 15.9 m.

The speed and height at 4 sec are 19.2 m/s and 1.6 m.

Answer from: Quest

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Answer from: Quest

The answer which is not a detail carson used to describe the pool in the cave at low tide is b. hundreds of periwinkles browse the branches and roots of the trees.

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