Question

a particle is dropped from a tower of height 2000m. find its velocity before it hits the ground and the time to reach the ground. (g=10m/s^2)

Answer 1

Answer:

Height from which the ball was dropped (h) = 2000m.

We know that velocity at ground is √2gh.

So, by this formula we have

or v = √2×10×2000

or v = √40000

Therefore v = 200m/s.

Also, time of descent =√2h/g

=√2×2000/10 = 20s.

Answer 2

Since, it is dropped from a tower, it will be dropped from rest, its initial velocity will 0 m/s.

u= 0 m/s

h = 2000 m

$g = 10 \: m \: {s}^{ - 2}$

From third equation of motion/gravitation, we conclude that

${v}^{2} = {u}^{2} + 2gh$

${v}^{2} = {0}^{2} + 2 \times 10 \: m {s}^{ - 2} \times 2000 \: m$

${v}^{2} = 40000 \: {m}^{2} \: {s}^{2}$

$v = \sqrt{40000 \: {m}^{2} \: {s}^{2} }$

$v = 200 \: m \: {s}^{ - 1}$

OR

v = 200 m / s