Question

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?

Answer 1

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 t:

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

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.

Answer 2

Answer:

Answer: Height is 0.204 m

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

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