Question

Ms. Burke is eating lunch on the flat roof of the science building. Her watermelon rolls to the edge moving directly horizontally at 6.18m/s and falls to the ground 16.6m below. How far from the side of the building does it land?

Answer 1

the watermelon lands approximately 11.38 meters from the side of the building.

Step-by-step explanation:

To find out how far from the side of the building the watermelon lands, we need to calculate the horizontal distance it travels before it hits the ground. We can do this by using the horizontal velocity and time of flight of the watermelon.

First, we need to calculate the time of flight. We can use the formula for vertical free fall:

h = 1/2 * g * t^2,

where h is the vertical distance (16.6 m) and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Plugging in the values, we can solve for t:

16.6 = 1/2 * 9.8 * t^2.

Simplifying the equation:

t^2 = (16.6 * 2) / 9.8,

t^2 ≈ 3.39,

t ≈ √3.39,

t ≈ 1.84 seconds.

Now that we have the time of flight, we can calculate the horizontal distance traveled. We use the formula:

d = v * t,

where d is the horizontal distance, v is the horizontal velocity (6.18 m/s), and t is the time of flight (1.84 seconds).

Plugging in the values:

d = 6.18 * 1.84,

d ≈ 11.38 meters.

Answer 2

11.369 meters

Step-by-step explanation:

To solve this problem, we can use the following steps:

1. Identify the known and unknown quantities.

Known quantities:

• Initial horizontal velocity, u = 6.18 m/s
• Distance fallen, h = 16.6 m

Unknown quantity:

• Horizontal distance traveled, d

2. Choose the appropriate equation of motion.

Since the watermelon is moving horizontally in free fall, we can use the following equation of motion:

`\boxed{\boxed{\sf d = u × t }}`

where:

• d is the horizontal distance traveled
• u is the initial horizontal velocity
• t is the time taken to fall

3. Solve for the unknown quantity.

We can rearrange the equation to solve for d:

`\boxed{\boxed{\sf d = u × t }}`

Substituting in the known values, we get:

`\sf d = 6.18 \: m/s\cdot t`

To find t, we can use the following equation of motion for free fall:

`\boxed{\boxed{\sf h = (1)/(2)\cdot g \cdot t^2 }}`

where:

• h is the distance fallen
• g is the acceleration due to gravity (9.81 m/s^2)
• t is the time taken to fall

Substituting in the known values, we get:

`\sf 16.6 m = (1)/(2)\cdot 9.81 m/s^2 \cdot t^2`

`\sf 16.6 = 4.905 t^2`

`\sf (16.6)/(4.905)=t^2`

`\sf 3.3843017329255 =t^2`

Solving for t, we get:

`\sf t = √( 3.3843017329255 )`

`\sf t \approx 1.8396471762067 s`

Substituting this value back into the equation for d, we get:

`\sf d = 6.18 \: m/s\cdot 1.8396471762067`

`\sf d = 11.369019548957`

`\sf d \approx 11.369 \textsf{( m in 3 d.p)}`

Therefore, the watermelon lands 11.369 meters from the side of the building.