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Ksju [112]
2 years ago
10

A grape is thrown straight up in the air. If it was thrown at 2 m/s, how high will it go in the air?

Physics
2 answers:
Elanso [62]2 years ago
6 0

Answer:

About 0.2041 meters.

Explanation:

We can use the following kinematic equation to determine the time in which the grape was in the air. (Ignoring air resistance.)

\displaystyle v_f = v_i + at

The final velocity is 0 m/s (the velocity at the peak is always zero), the initial velocity is 2 m/s, and the acceleration (due to gravity) is -9.8 m/s². Solve for time <em>t: </em>

<em />\displaystyle \begin{aligned} (0\text{ m/s}) & = (2 \text{ m/s}) + (-9.8 \text{ m/s$^2$})t \\ \\ t & =\frac{-2\text{ m/s}}{-9.8 \text{ m/s$^2$}} \\ \\ & =0.2041 \text{ s} \end{aligned}<em />

<em />

To find the distance traveled, we can another kinematic equation:

\displaystyle \Delta d = v_i t + \frac{1}{2} a t^2

The initial velocity is 2 m/s, the acceleration (due to gravity) is -9/8 m/s², and the time it took for the graph to reach its maximum height is 0.2041 seconds. Hence:

\displaystyle \begin{aligned} \Delta d & = (2\text{ m/s})(0.2041 \text{ s}) + \frac{1}{2}(-9.8 \text{ m/s$^2$})(0.2041\text{ s})^2 \\ \\ & = 0.2041 \text{ m}  \end{aligned}

In conclusion, the grape travels 0.2041 meters in the air.

ser-zykov [4K]2 years ago
4 0

Answer:

Answer: <u>Height</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>2</u><u>0</u><u>4</u><u> </u><u>m</u>

Explanation:

At the highest point, it is called the maximum height.

• From third newton's equation of motion:

{ \rm{ {v}^{2}  =  {u}^{2}  + 2gs}}

• At maximum height, v is zero

• u is initial speed

• g is -9.8 m/s²

• s is the height

{ \rm{0 {}^{2}  =  {2}^{2}   -  (2 \times 9.8 \times s)}} \\  \\ { \rm{4 = 19.6s}} \\  \\ { \rm{s = 0.204 \: m}}

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The time taken by the stone to hit the ground would be 5.12 seconds.

<h3>What are the three equations of motion?</h3>

There are three equations of motion given by  Newton

The first equation is given as follows

v = u + at

the second equation is given as follows

S = ut + 1/2×a×t²

the third equation is given as follows

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Keep in mind that these calculations only apply to uniform acceleration.

As given in the problem, a stone is dropped from the helicopter which is ascending at the speed of 19.6 m/s

height(S) = 156.8 meters

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Learn more about equations of motion from here,

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