Physics Help. A rescue plane flying horizontally at 250 km/hr wants to drop supplies to stranded hikers...?

(a)

Let:

x be the horizontal distance travelled by the supplies before they reach the ledge,

y be the vertical descent to the ledge,

u be the horizontal speed of the plane,

t be the time for the supplies to fall,

g be the acceleration due to gravity.

x = ut

y = gt^2 / 2

250 km / hr = 2.5 * 10^5 / 3600 = 69.444 m/s.

Eliminating t:

x^2 = 2u^2 y / g

= 2 * 69.444^2 * 200 / 9.81

x = 443.4 m.

(b)

Let:

u1 be the horizontal speed of the plane,

u2 be the initial vertical downward speed given to the supplies.

x = u1 t

y = u2 t + gt^2 / 2

Eliminating t:

y = u2(x / u1) + (g / 2)(x / u1)^2

u2 = [ y - (g / 2)(x / u1)^2 ] / (x / u1)

= [ 200 - (9.81 / 2)(400 / 69.444)^2 ] / (400 / 69.444)

= 6.47 m/s.

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