Question

You throw a rock straight up and find that it returns to your hand 3.40 s after it left your hand. Neglect air resistance. What was the maximum height above your hand that the rock reached?

Answer 1

Answer:

The maximum height of the rock is 14.2 m

Explanation:

The equations that describe the height and velocity of the rock are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the object at time t

y0 = initial height

t = time

g = acceleration due to gravity (-9.8 m/s² if upward is positive)

v = velocity of the object at time t

We know that at t = 3.40 s, the rock is in your hand again. Then, if we place the origin of the frame of reference at your hand, the position of the rock at 3.40 s is 0 m. Using the equation of the position, we can calculate the initial velocity that we will need to obtain the max-height.

y = y0 + v0 · t + 1/2 · g · t²

0 = v0 · 3.40 s - 1/2 · 9.8 m/s² · (3.40 s)²

(1/2 · 9.8 m/s² · (3.40 s)² ) / 3.40 s = v0

v0 = 16.7 m/s

At max-height, the velocity of the rock is 0. Then, using the equation of velocity we can calculate the time it takes the rock to reach the max-height. With that time, we can calculate the maximum height.

v = v0 + g · t      (at max-height, v = 0)

0 = 16.7 m/s - 9.8 m/s² · t

- 16.7 m/s /  - 9.8 m/s² = t

t = 1.70 s

Now, using this time in the equation of height:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 16.7 m/s · 1.70 s - 1/2 · 9.8 m/s² · (1.70 s)²

y = 14.2 m

The maximum height of the rock is 14.2 m

Answer 2

The maximum height above your hand that the rock reached is equal to 14.16 meters.

Given the following data:

• Time = 3.40 seconds

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 $m/s^2$.

To determine the maximum height above your hand that the rock reached:

First of all, we would determine the initial velocity of the rock before it reaches its maximum height by using the second equation of motion:

Mathematically, the second equation of motion is given by the formula;

$S=ut -\frac{1}{2} at^2$

Where:

• S is the distance covered.

• u is the initial velocity.

• a is the acceleration.

• t is the time measured in seconds.

Substituting the given parameters into the formula, we have;

$0 = u(3.40) - \frac{1}{2} \times 9.8 \times 3.40^2\\\\3.40u = 4.9 \times 11.56\\\\3.40u = 56.644\\\\u = \frac{56.644}{3.40}$

Initial velocity, u = 16.66 m/s

Next, we would determine the time required for the object to reach maximum height by using the first equation of motion:

$V = u - at\\\\0 = 16.66 - 9.8t\\\\9.8t = 16.66\\\\t = \frac{16.66}{9.8}$

Time, t = 1,7 seconds

Mathematically, maximum height is given by the formula:

$H = \frac{1}{2} gt^2\\\\H = \frac{1}{2} \times 9.8 \times 1.7^2\\\\H = 4.9 \times 2.89$

Maximum height, H = 14.16 meters

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