Question

a body falling freely from a tower of height 100m, after 50sec, it reaches the ground, calculate it's final velocity just before it touches the ground​

Answer 1

Answer:

4m/s

Explanation:

by 2nd equation of motion, S= ut + 1/2at^2

we can find the acceleration as follows -->

$100=0.t + \frac{1}{2}.a.50^{2}$

100= a.2500/2

100=a.1250

therefore a=2/25  m/s^2

hence by first equation of motion,

v=u+at

v= 0 + 2/25 X 50

v=4m/s

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Answer 2

Answer:

The final velocity of the freely falling body will be 4 m/s.

Explanation:

The distance traveled by the body, S = 100 m

Time taken to reach this distance, t = 50 s

The body is freely falling. So its initial velocity, u = 0 m/s

The acceleration can be found using the 2nd equation of motion.

S = ut + (at²/2)

⇒ a = (S - ut) 2/t²

   a = (100 - 0) 2/50²

   a = 200 / 2500

   a = 0.08 m/s²

Now the final velocity can be found using the 1st equation of motion.

v = u + at

v = 0 + (0.08 × 50)

v = 4 m/s