The pilot of an airplane traveling at 180 km/hwants to drop supplies to flood vi

Question

The pilot of an airplane traveling at 180 km/hwants to drop supplies to flood victims isolated on a patch of land99 m below. How long should the pilot drop the supplies before theplane is directly overhead? (Assume no airresistance.)

What is the horizontal distance from the plane tothe victims when the supplies are dropped?


Someone said to use this: not sure though : I plugged it in butdidn't get the right answer online...........

use y=y0 - .5*g*t^2

y=0 b/c its the ground, y0 is height of the plane. solve for
t, and thats the first answer.

For the second one use

x=x0 + Vx0 * T
Plug in the T, Vx0 is given but convert it to m/s.

The pilot of an airplane traveling at 180 km/hwants to drop supplies to flood victims isolated on a patch of land99 m below. How long should the pilot drop the supplies before theplane is directly overhead? (Assume no airresistance.)

What is the horizontal distance from the plane tothe victims when the supplies are dropped?


Someone said to use this: not sure though : I plugged it in butdidn't get the right answer online...........

use y=y0 - .5*g*t^2

y=0 b/c its the ground, y0 is height of the plane. solve for
t, and thats the first answer.

For the second one use

x=x0 + Vx0 * T
Plug in the T, Vx0 is given but convert it to m/s.

Explanation / Answer

Let the time interval be t. If the speed of plane is v, use vt = x to get thehorizontal distance travelled by the plane. Since the package was moving horizontally at the samespeed, there being no horizontal acceleration, its speed remainsthe same. So, horizontal distance covered by package = vt =x Now, in the same time t, the package must have reachedthe people below vertically that is the ground level. Use y = Uy*t + a*t*t/2 if you take downward as positive, y = height of plane = 0*t + 9.8*t*t / 2 (sincethe initial vertical speed was zero) So, t = (2y/9.8) = 4.495 s So, x = v*t = 180*(1000 m/km) / (3600 s/hr) * t = 224.745 m

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