Question

2. an object dropped from cliff, acceleration is constant @10 m/s^2. find the speed 5sec after it was dropped.​

Answer 1

Given :

• Acceleration of the object = 10 m/s²
• Time = 5 sec

To Find :

• The speed of the dropped object after 5 sec

Solution :

From first equation of motion ,

$\\ \star \: {\boxed{{\sf{ \purple{v = u + at}}}}}$

Where ,

• v is final velocity
• u is initial velocity
• a is acceleration
• t is time

We have ,

• u = 0 [Since the object is dropped]
• v = v (let)
• t = 5 s
• a = 10 m/s²

Substituting the values ,

$\\ : \implies \sf \: v = 0 + (10)(5)$

$\\ : \implies \sf \: v =0 + 50 \: m {s}^{ - 1}$

$\\ : \implies{\underline{{\boxed{\pink{\mathfrak{v = 50 \: m{s}^{ - 1} }}}}}} \: \bigstar$

Hence ,

• The Speed of the object after 5 sec it was dropped is 50 m/s.

Answer 2

Given:-

•An object dropped from cliff , acceleration is constant at 10 m/s².

To Find:-

•The speed of the object (dropped) after 5 second

Solution:-

•Here, they asked velocity of the object dropped after 5 seconds..By using First equation of motion, we can find the velocity of the dropped object....

•We know,

a = 10 m/s²

t = 5 seconds

u = 0

v = ?

•Substitute the given in the formula,

v = u + at

•Let v be 'x'

x = 0 × 10 × 5

x = 50

Hence,

$\bigstar { \underline{ \boxed{ \sf{v \: = 50 \: m {s}^{ - 1} }}}} \bigstar$