Physics Question??

Question

rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge H = 221 m below. If the plane is traveling horizontally with a speed of 255 km/hr (70.8 m/s), how far in advance of the recipients (horizontal distance) must the goods be dropped?

Suppose, instead, that the plane releases the supplies a horizontal distance of x = 401 m in advance of the mountain climbers. What vertical velocity (use the positive direction as upwards) should the supplies be given so that they arrive precisely at the climbers' position?

With what speed do the supplies land in the latter case?

I know the answer to the first part is 476 m, but how do you go about solving the last two questions?

Answer 1

is this like, a question for school or something?

Answer 2

Horizontal distance Problem:

First solve for the time it takes the supplies to reach the recipients using the formula:

s=ut+1/2at^2

where u is the initial vertical component of the velocity; in this case u is 0 because the plane is traveling horizontally. It's like the supplies are dropped and they fall freely to a height s of 221 meters. Now substitute
the given values:

221=o*t+1/2*9.8t^2
221=4.9t^2
t^2=221/4.9
=45.1
t=45.1^1/2
=6.72sec

Then use the formula Distance=Velocity*Time
where Distance is the horizontal distance, Velocity =70.8m/s, and Time=6.72 sec.

Substitute the given values:

Distance=70.8*6.72=475.8meters

Vertical Velocity Problem:

In this case the Distance=401meters, the horizontal
velocity is still the same=70.8m/s. So we calculate the
time needed for the drop. Substitute the given values in the formula Distance=Velocity*Time

401=70.8t
t=401/70.8
=5.66sec

Now use the formula s=ut+1/2at^2

where s=221, u =the unknown vertical velocity, and
t=5.66sec. Substitute the given values in the formula:

221=u*5.66+1/2*9.8*5.66^2
221=5.66u+156.8
5.66u=156.8-221
=-64.2
u=-64/5.66
=-11.3m/s

If upwards is positive then negative means downwards, meaning the supplies have to be given a downward push of 11.3m/s to reach the climbers.

Final Velocity Problem:

Use the formula, v=u+at, where v is the final velocity,
a =-9.8m/s^2, and t=5.66sec. Substitute the given
values in the formula:

v=-11.3+(-9.8)*5.66
=-11.3-55.5
=-66.8m/s also negative, meaning the direction is
downwards(of course).

Answer 3

You know your kinematics equations, right? You can use the final x position to solve for the amount of time the supplies spends in the air. Use the time to solve for the vertical displacement, using your airplane's height, acceleration due to gravity, and initial velocity as the variable.

For the last part, just use the time it takes for the supplies to land and the initial velocity to solve for the vertical velocity. The horizontal velocity is just the velocity of the plane. Combine the two with pythagorean theorem to get the total speed.

Answer 4

Calculate the time it takes to travel the horizontal distance, plug that into -221m = v*t-1/2gt^2 and solve for v, using g>0

Then vertical vfinal=v-1/2gt^2 for