A package is dropped from a helicopter flying at a height of 7.5 m while traveling 15 m/s. How far is the package travel horizontally before it hits the ground below?

h = 0.5g*t^2 = 7.5

4.9t^2 = 7.5
t^2 = 1.53
t = 1.24 s. = Fall time.

d = Vx*t = 15m/s * 1.24s = 18.6 m.

h = 0.5g*t^2 = 7.5

4.9t^2 = 7.5
t^2 = 1.53
t = 1.24 s. = Fall time.

d = Vx*t = 15m/s * 1.24s = 18.6 m.

1 0

Well, let's calculate it in a fun way, shall we? Picture this: the package is feeling pretty adventurous, so it decides to take a little leap from the helicopter. But don't worry, it doesn't scream like a roller coaster rider!

Now, to determine the horizontal distance the package travels, we need to consider two things: the time it takes to hit the ground and its initial horizontal velocity. Since the package is dropped vertically, we know it doesn't have any initial horizontal velocity. It's like a penguin belly-flopping into the water!

Now, let's find the time it takes for the package to hit the ground. We can use the equation:

d = (1/2) * g * t^2,

where d is the vertical distance (7.5m in this case), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

If we rearrange the equation to solve for t, we get:

t = sqrt((2 * d) / g).

Now, let's plug in the numbers:

t = sqrt((2 * 7.5) / 9.8) ≈ 0.973 seconds.

Once the package starts its thrilling freefall journey, it will take approximately 0.973 seconds to hit the ground. During this time, its horizontal velocity doesn't change, remaining at 15 m/s.

So, the horizontal distance the package travels before it hits the ground can be calculated using:

distance = velocity * time.

Plugging in the values:

distance = 15 * 0.973 ≈ 14.6 meters.

Therefore, the package will travel approximately 14.6 meters horizontally before it reaches the ground. Who knew package-delivery could be so exciting?

To find the horizontal distance traveled by the package before it hits the ground, we can use the formula for horizontal distance traveled:

Distance = velocity × time

Since the package is dropped from a helicopter, its initial horizontal velocity is assumed to be zero. Therefore, the only force acting on the package is gravity, which causes it to accelerate vertically downward. We can find the time it takes for the package to hit the ground using the formula:

Distance = initial velocity × time + (1/2) × acceleration × time^2

Since the initial vertical velocity of the package is zero, the formula simplifies to:

Distance = (1/2) × acceleration × time^2

The acceleration due to gravity is approximately 9.8 m/s^2. We can rearrange the formula to solve for time:

time = √(2 × distance / acceleration)

Substituting the given values, we get:

time = √(2 × 7.5 m / 9.8 m/s^2)
≈ √(1.53 s)
≈ 1.24 s

Now, we can find the horizontal distance traveled using the formula for horizontal distance:

Distance = velocity × time

Substituting the given values, we get:

Distance = 15 m/s × 1.24 s
≈ 18.6 m

Therefore, the package travels approximately 18.6 meters horizontally before it hits the ground.

To find the horizontal distance traveled by the package before hitting the ground, we can use the formula for horizontal distance:

d = v * t

where
d is the horizontal distance traveled,
v is the horizontal velocity of the package, and
t is the time it takes for the package to hit the ground.

We need to find the time it takes for the package to hit the ground. Since the package is dropped from rest, we can use the equation for vertical motion:

h = (1/2) * g * t^2 + v_i * t + h_i

where
h is the vertical distance dropped (7.5 m),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time it takes for the package to hit the ground, and
v_i and h_i are the initial velocity and initial height, respectively.

Since the package is dropped from rest, the initial vertical velocity (v_i) is 0 m/s, and the initial height (h_i) is 0 m.

Substituting these values into the equation, we have:

7.5 m = (1/2) * 9.8 m/s^2 * t^2 + 0 m/s * t + 0 m

Simplifying the equation, we get:

7.5 m = 4.9 m/s^2 * t^2

Dividing both sides of the equation by 4.9 m/s^2, we get:

1.53 s^2 = t^2

Taking the square root of both sides, we find:

t ≈ 1.24 s

Now that we have the time it takes for the package to hit the ground, we can find the horizontal distance traveled:

d = v * t
d = 15 m/s * 1.24 s
d ≈ 18.6 m

Therefore, the package will travel approximately 18.6 meters horizontally before hitting the ground below.