Physics, 02.07.2019 23:40 econsta3

Astone is dropped from the upper observation deck of a tower, 300 m above the ground. (assume g = 9.8 m/s2.) (a) find the distance (in meters) of the stone above ground level at time t. h(t) = correct: your answer is correct. (b) how long does it take the stone to reach the ground? (round your answer to two decimal places.) 7.82 correct: your answer is correct. s (c) with what velocity does it strike the ground? (round your answer to one decimal place.) -76.64 correct: your answer is correct. m/s (d) if the stone is thrown downward with a speed of 2 m/s, how long does it take to reach the ground? (round your answer to two decimal places.) 8.00 incorrect: your answer is incorrect. s

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Answer from: james169196

Answers:

(a) $y(t)=300m-4.9m/s^{2}t^{2}$

(b) t=7.82s

(d) t=7.6s

Explanation:

This problem is a good example of vertical motion, and the main equations for this situation are as follows:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

$V=V_{o}-gt$ (2)

Where:

is the height of the stone at a given time

$y_{o}=300m$ is the initial height of the stone

$V_{o}$ is the initial velocity of the stone

is the time

$g=9.8m/s^{2}$ is the acceleration due to gravity on Earth

Now, for parts (a), (b) and (c) we are specifically talking about free fall (where the main condition is that the initial velocity must be zero $V_{o}=0$ ).

How do we know this?

Because we are told "the stone is dropped".

Having this clear, let's start with the calculations:

(a) Find the distance of the stone above ground level at time t:

In order to approach this, we will use equation (1), remembering the condition $V_{o}=0$ :

$y(t)=y_{o}-\frac{1}{2}gt^{2}$ (3)

$y(t)=300m-\frac{1}{2}(9.8m/s^{2})t^{2}$ (4)

$y(t)=300m-4.9m/s^{2}t^{2}$ (5) This is the distance of the stone above ground level at time t

(b) How long does it take the stone to reach the ground?:

In this case, y=0 , because we have to find the time when the stone finally hits the ground.

Rewritting (1) with this condition:

$0=y_{o}-\frac{1}{2}gt^{2}$ (6)

Isolating :

$t=\sqrt{\frac{2y_{o}}{g}}$ (7)

$t=\sqrt{\frac{2(300m)}{9.8m/s^{2}}}$ (8)

Then:

t=7.82s (9)

In this part, rewritting equation (2) will be usefull, remembering $V_{o}=0$ and using the timewe found in (9), which is the time when the stone strikes the ground:

V=-gt (10)

$V=-(9.8m/s^{2})(7.82s)$ (11)

$V=-76.68m/s \approx -76.7 m/s$ (12) Note the velocity has a negative sign because its direction is downwards.

(d) Finding

when $y_{o}=0$ and $V_{o}\neq 0$ :

Now the initial velocity is $V_{o}=-2m/s$ (downwards). So, let's use equation (1) again an find :

$0=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (13)

$0=300m+(-2m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$

$0=-4.9m/s^{2}t^{2}+(-2m/s)t+300m$ (14)

At this point we have a quadratic equation of the form $0=at^{2}+bt+c$ , and we have to use the quadratic formula if we want to find :

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ (15)

Where a=-4.9 , b=-2 , c=300

Substituting the known values and choosing the positive result of the equation:

$t=7.623 s \approx 7.6 s$ This is the time it takes to the stone to reach the ground when $V_{o}=-2m/s$

Answer from: Quest

Is there a picture to go along with this

Answer from: Quest

Sorry my friend i don't know the answer. i am just in 6th
Me : ) it doesn't need to be a paragraph just a list

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Astone is dropped from the upper observation deck of a tower, 300 m above the ground. (assume g = 9....

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