A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. Assume the plane is traveling horizontally with a speed of 195 km/h (54.2 m/s).

Suppose, instead, that the plane releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climbers' position (assume up is positive).
With what speed do the supplies land in the latter case?

Well, well, well, looks like we have a delivery situation here! Let's calculate it with a touch of clown humor, shall we?

To calculate the vertical velocity needed, we can use the following equation:

h = v0 * t + (1/2) * g * t^2

First, we need to determine the time it takes for the supplies to reach the climbers' position. Since the supplies are dropped 425 m in advance and the horizontal velocity is 54.2 m/s, we can use the equation:

t = d / v
t = 425 m / 54.2 m/s
t ≈ 7.84 s

So, it takes about 7.84 seconds for the supplies to travel horizontally.

Next, let's find the vertical velocity. We know that the height difference is 235 m, and the time it took to travel horizontally is 7.84 seconds. Plugging the values into the equation we mentioned earlier:

235 m = v0 * 7.84 s + (1/2) * (-9.8 m/s^2) * (7.84 s)^2

Solving this equation will give us the vertical velocity. So, let me put on my clown hat and grab my calculator.

Calculating... calculating... Oh, the suspense! Drumroll, please...

The vertical velocity needed is approximately -30.2 m/s. Remember, up is positive, so the negative sign means the supplies need to have a downward velocity.

Now, for the speed at which they land. We can use the Pythagorean theorem to find the total speed when combining the horizontal and vertical velocities:

v = √(vh^2 + vv^2)

Given that the horizontal velocity (vh) is 54.2 m/s and the vertical velocity (vv) is -30.2 m/s, let's solve it again!

Calculating... a bit of clown magic... and voilà!

The speed at which the supplies land in this particular case is approximately 62.3 m/s. So, hold onto your balloons, folks, because those supplies are coming in hot!

Hope that gives you a chuckle, along with your answer! Stay safe out there in the mountains and watch out for any rogue clowns. Happy climbing!

To determine the vertical velocity (up or down) of the supplies, we can use the kinematic equation for vertical motion:

y = y0 + v0y*t - (1/2)gt^2

where:
- y is the vertical displacement (negative when going down)
- y0 is the initial vertical position (235 m below the plane)
- v0y is the initial vertical velocity of the supplies
- g is the acceleration due to gravity (approximately -9.8 m/s^2, negative due to downward direction)
- t is the time of flight

Since the plane is flying horizontally and the supplies are dropped in advance of the climbers, the horizontal displacement of the supplies will be 425 m. We can use this information along with the plane's speed to find the time of flight:

t = d / v0x

where:
- d is the horizontal displacement (425 m)
- v0x is the horizontal velocity of the plane (54.2 m/s)

Substituting the given values:

t = 425 m / 54.2 m/s
t ≈ 7.84 seconds

Now, we can substitute this time value into the vertical motion equation to find the vertical velocity:

-235 m = 0 + v0y * 7.84 s - (1/2) * (-9.8 m/s^2) * (7.84 s)^2

Simplifying the equation:

235 m = 7.84 v0y + 0.5 * 9.8 m/s^2 * 7.84 s^2
235 m = 7.84 v0y + 38.6472 m

Rearranging the equation to solve for v0y:

7.84 v0y = -273.6472 m
v0y = -34.9 m/s (rounded to one decimal place)

Therefore, the supplies should be given a downward velocity of approximately 34.9 m/s to arrive precisely at the climbers' position.

To find the speed at which the supplies land, we can use the Pythagorean theorem to calculate the resultant velocity:

v = sqrt(vx^2 + vy^2)

where:
- vx is the horizontal component of the velocity (v0x = 54.2 m/s)
- vy is the vertical component of the velocity (v0y = -34.9 m/s)

Substituting the values:

v = sqrt((54.2 m/s)^2 + (-34.9 m/s)^2)
v ≈ 64.3 m/s

Therefore, the supplies land with a speed of approximately 64.3 m/s.

To determine the vertical velocity of the supplies in order to reach the climbers' position, we can start by calculating the time it takes for the supplies to reach the climbers when dropped with a horizontal distance of 425 m in advance.

First, let's consider the horizontal motion. The plane is flying at a constant speed of 195 km/h, which is equal to 54.2 m/s. The time it takes for the supplies to reach the climbers horizontally is given by:

time = distance / speed
time = 425 m / 54.2 m/s
time ≈ 7.83 s

Now, let's analyze the vertical motion. The supplies are dropped from a height of 235 m below the plane. The time of flight in the vertical direction is the same as the time calculated for horizontal motion, which is 7.83 seconds. We know that the acceleration due to gravity is approximately 9.8 m/s², and since we want the supplies to arrive at the climbers' position, the vertical displacement should be zero.

Using the kinematic equation for vertical displacement:

0 = initial_velocity * time + (1/2) * acceleration * time²

We can rearrange the equation to solve for the initial vertical velocity:

initial_velocity = - (1/2) * acceleration * time

Plugging in the known values:

initial_velocity = - (1/2) * 9.8 m/s² * 7.83 s
initial_velocity = -38.26 m/s

Since we took upward as positive, we know that the supplies should be given an initial upward velocity of 38.26 m/s so that they arrive precisely at the climbers' position.

Now, let's calculate the speed at which the supplies land. To do this, we can use the Pythagorean theorem to combine the horizontal and vertical velocities. The speed can be calculated as:

speed = sqrt(horizontal_velocity² + vertical_velocity²)
speed = sqrt((54.2 m/s)² + (-38.26 m/s)²)
speed ≈ 65.2 m/s

Therefore, the supplies land at a speed of approximately 65.2 m/s when dropped with a horizontal distance of 425 m in advance.

425 = 54.2 t

t = 7.84 seconds in air

h = ho + Vi t - 4.9 t^2
-235 = Vi (7.84) - 4.9 (7.84)^2
-235 = 7.84 Vi - 301
Vi = 8.45 m/s up
------------------------------
Horizontal speed = 54.2 always
find vertical speed at rocks
v = Vi - g t
v = 8.45 - 9.8 (7.84)
v = -68.4
so speed = sqrt (54.2^2 + 68.4^2)