Question

An object was dropped from cliff falls with a constant acceleration of 10 m/s2 . Find it's speed 5s after it was dropped

Answer 1

Explanation:

$\sf{Given}\begin{cases}\sf{u=0 ms^{-1}}\\ \sf{a=-10 ms^{-2}}\\ \sf{t= 5s}\\ \sf{v=??}\end{cases}$

★Using,,

$\sf{\boxed{v=u+at}}$

★We get,,

$\sf{v=0+(-10)(5)}$

$\sf{\implies{v=-50 ms^{-1}}}$

★Hence,,

Speed of the object after 5s =-50 m/s

( Here -ve sign for downward direction )

$\rule{200}2$

Hope it helps ❣️❣️❣️

Answer 2

Answer:

50 m/s

Explanation:

V = u + at

Initial velocity or u = 0

Acceleration or a = 10 m/s²

Time of journey or t = 5 seconds

So,

V = 0 + 10 x 5

V = 10 x 5 = 50

Speed:

The Speed of a body is the rate of change of distance with time.

Numerically it is the distance travelled by the body in one second.

It is a scalar quantity.

It is generally represented by the letter u or v.