Question

A body is thrown upwards with 10 m/s. There's always an acceleration
of 10m/s
2
in a downward direction. What is the time taken by the body

to come back?

Answer 1

Given:-

• Initial velocity ,u = 10m/s
• Final velocity ,v = 0m/s
• Acceleration ,a = -10m/s²

To Find:-

• Time taken by the body to come back,t

Solution:-

According to the Question

we have to calculate the total time taken by the body to come back . So , we apply here first equation of motion

• v = u + at

where,

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

substitute the value we get

$\longrightarrow$ 0 = 10 + (-10) × t

$\longrightarrow$ 0 -10 = -10×t

$\longrightarrow$ -10 = -10×t

$\longrightarrow$ 10 = 10×t

$\longrightarrow$ 10 × t = 10

$\longrightarrow$ t = 10/10

$\longrightarrow$ t = 1s

• Hence, the time taken by the body to come back is 1 second.

Answer 2

Answer:

Given :-

• A body is thrown upwards with 10 m/s. There's always an acceleration of 10 m/s² in a downward direction.

To Find :-

• What is the time taken by the body to come back.

Formula Used :-

$\clubsuit$ First Equation Of Motion :

$\mapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time

Solution :-

Given :

$\bigstar\: \: \rm{\bold{Final\: Velocity\: (v) =\: 0\: m/s}}$

$\bigstar\: \: \rm{\bold{Initial\: Velocity\: (u) =\: 10\: m/s}}$

$\bigstar\: \: \rm{\bold{Acceleration\: (a) =\: - 10\: m/s^2}}$

According to the question by using the formula we get,

$\longrightarrow \sf 0 =\: 10 + (- 10)t$

$\longrightarrow \sf 0 =\: 10 - 10t$

$\longrightarrow \sf 0 - 10 =\: - 10t$

$\longrightarrow \sf {\cancel{-}} 10 =\: {\cancel{-}} 10t$

$\longrightarrow \sf 10 =\: 10t$

$\longrightarrow \sf \dfrac{1\cancel{0}}{1\cancel{0}} =\: t$

$\longrightarrow \sf \dfrac{1}{1} =\: t$

$\longrightarrow \sf 1 =\: t$

$\longrightarrow \sf\bold{\red{t =\: 10\: seconds}}$

$\therefore$ The time taken by the body to come back is 1 seconds.