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# Abody falling freely under gravity passed two points 30 m apart in 1s from what point above the upper point it began to fall

• Réponse publiée par: lhadyclaire

Step-by-step explanation:

firstly we have to find the v

so

v=u+at

v= 0+(9.8)(1)

v=9.8m/s

so for distance

2as=v^2 -u^2

2(9.8)(s) = 9.8^2

2×9.8(s) = 9.8×9.8

s= 9.8/2

s= 4.9 metre

• Réponse publiée par: RoseTheShadowHunter

S = v t + 1/2 g t^2      where S = 30 m   and t = 1 s

30 = v * 1 + (9.8 / 2) * 1

v = 30 - 4.9 = 25.1

So V = 25.1 m/s   when it passed the upper point

V = g T   where T is the time elapsed to reach the upper point

T = 25.1 / 9.8 = 2.56 s

H = 1/2 g t^2 = (9.8 / 2) * 2.56^2    where H is height above upper point

H = 32.1 m

As a check:

Vu = 9.8 * 2.56 = 25.1 m/s    speed at upper point

Vl = 9.8 * 3.56 = 34.9 m/s     speed at lower point

2 g S = 34.9^2 - 25.1^2 = 588

S = 588 / 19.6 = 30 m

• Réponse publiée par: homersoncanceranguiu

microwave = ooh a pizza, yummy

infrared = the eyes of snakes, nani?

visible light = hello there fellow human

ultraviolet = used to detect fake documents

x-ray = wa! a skeleton

gamma ray = shiboom that's the last nuclear
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Abody falling freely under gravity passed two points 30 m apart in 1s from what point above the uppe...