Question

Small-plane pilots regularly compete in "message drop" competitions, dropping heavy weights (for which air resistance can be ignored) from their low-flying planes and scoring points for having the weights land close to a target. a plane 70 m above the ground is flying directly toward a target at 44 m/s . at what distance from the target should the pilot drop the weight?

Answer 1

Answer:

1.7 × 10² m

Explanation:

The movement of the weight can be decomposed in a vertical component and a horizontal component.

The vertical movement is uniformly accelerated motion (constant acceleration) and is the one that we will use to find the time of flight (t). The initial vertical speed is zero, and the vertical distance (y) traveled is 70 m. The acceleration is that of gravity.

y = 1/2 . a. t²

t = √(2y/a) = √(2 . 70 m/ 9.8 m/s²) = 3.8 s

The horizontal movement is a uniform motion (constant speed). The  horizontal speed is that of the plane. The horizontal distance at what the pilot should drop the weight is:

d = v . t = 44 m/s . 3.8s = 1.7 × 10² m

Answer 2

plane is flying at an altitude of 70 m

now if an object is dropped from it then time taken by object to drop on ground will be given as

$y = v_i* t + \frac{1}{2}at^2$

here initial speed in vertical direction must be zero as plane is moving horizontal

given that

y = 70 m

a = 9.8 m/s^2

$70 = 0 + \frac{1}{2}*9.8*t^2$

$t = 3.77 s$

now since the plane is moving horizontally with speed v = 44 m/s

so the horizontal distance moved by the object will be

$d = v_x * t$

$d = 44 * 3.77$

$d = 166.3 m$

so the distance moved by the box is 166.3 m