A lead weight starts from rest at time t=0 and falls freely under gravity..?

Question

A lead weight starts from rest at time t=0 and falls freely under gravity. Calculate its velocity, distance moved and acceleration towards the ground after 5 seconds

v(final) = v(initial) + at
v(final) = 0 + 10 x 5
v(final) = 50 ms-1 downwards?
but i got it wrong!! plz help??

Answer 1

Whats the answer?

Answer 2

v = u + at where a = g = 9.81 m/sec^2

v = 0 + 5g = 5 x 9.81 = 49.5 m/sec after 5 secs. If you take g as 10m/sec^2 v = 50 m/sec (downwards).

Acceleration = g = 9.81m/sec^2 (or 10m/sec^2 approx.)

Distance covered in 5 secs, s is given by

s = ut + 1/2 at^2 = 0.5 + 1/2.g.25 = 25/2 x 9.81 = 122.625 m

= 25/2 x 100 = 125 m if we take g as 10 m/sec^2

Answer 3

If it falls under gravity from rest then:

u(initial) = 0m/s
v(final) = ?
t = 5s
a(acceleration) = 9.81m/s2 (acceleration due to gravity)
s (distance) = ?

So we already found acceleration which is constant throughout

We can find velocity next as a = v-u/t

v = at - u
v =(9.81 x 5) - 0
v = 49.1m/s

now we can use v^2 = u^2 +2as to find s
s =( v^2 - u^2)/2a
s = (49.05^2 - 0^2)/(2 x 9.81)
s = 122.6m travelled, hence it was released 122.6m above ground.

Answer 4

v= 240

9.6 meters per sec^2

9.6*5^2

9.6*25

240 meters/sec = v(final)