There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. A pilot flying low and slow drops a weight; it takes 2.0 s to hit the ground, during which it travels a horizontal distance of 100 m. Now the pilot does a run at the same height but twice the speed. How much time does it take the weight to hit the ground? How far does it travel before it lands?

Since the plane is going twice as fast, the weight will travel twice as far (ignoring air resistance). So it will travel a horizontal distance of 200 m.

Answer Link

dotmanjohn dotmanjohn · Answer 2 · 2020-01-18T10:37:58+01:00

Answer:

1) 2 seconds

2) 200 m

Explanation:

1) Fall time at initial speed [tex]s_{1}[/tex] = [tex]t_{1}[/tex]

Fall time at final speed [tex]s_{2}[/tex] = [tex]t_{2}[/tex]

Initial fall height [tex]h_{1}[/tex] at initial speed = Final fall height [tex]h_{2}[/tex] at final speed i.e [tex]h_{1}[/tex] = [tex]h_{2}[/tex]

s = speed

t = time

h = height

Therefore, since fall time depends on fall height where acceleration due to gravity (g) is constant,

Fall time at [tex]s_{1}[/tex] = Fall time at [tex]s_{2}[/tex]

i.e [tex]t_{1}[/tex] = [tex]t_{2}[/tex] = 2.0 s

Time taken to land = 2.0 s

2) Initial distance traveled ([tex]S_{1}[/tex]) at initial speed [tex]s_{1}[/tex] = 100 m

Final speed [tex]s_{2}[/tex] is double initial speed i.e [tex]s_{2}[/tex] = [tex]2s_{1}[/tex]

Therefore, since distance traveled is directly proportional to speed,

Final distance traveled [tex]S_{2}[/tex] at final speed [tex]s_{2}[/tex] is double initial distance [tex]S_{1}[/tex]

i.e [tex]S_{2}[/tex] = [tex]2S_{1}[/tex]

[tex]2S_{1}[/tex] = 2 x 100 m = 200 m

Distance traveled = 200 m