Question

a particle is projected vertically upwards from the ground with an initial velocity of u m/s ,the time of flight is
a) u/g
b) u2/g
c) 2u/g
d)none​

Answer 1

Answer:

2u/g

Explanation:

Using the equations of motion:

 v = u + at

At the highest point, v = 0  , a = -g

⇒ v = u + at

⇒ 0 = u + (-g)t

⇒ - u = - gt

⇒ u/g = t

It is the time taken when particle is thrown upwards and reached the highest time.

The same time it would take to come back.

Total time = u/g + u/g = 2u/g

Answer 2

Required Answer :-

$\sf{ \pink{ \star{Given}{} }{} }{}$

• The Particle projected upward has v = 0
• a = -g ( ball thrown against the Gravity )

$\sf{ \pink{ \star{To \: Find}{} }{} }{}$

• The time of flight.

$\sf{ \pink{ \star{Solution}{} }{} }{}$

We Are needed to Find the time taken by the particle

When a particle moves up at the highest point the value of final velocity v is zero 0 and the initial velocity u m/s and acceleration -g which is against gravity.

We Use the first Equation of Motion to derive

$\sf{ \pink{ \star{v = u + at}{} }{} }{}$

Let us substitute the Values,

$\sf{ \rightarrow{v = u + at}{} }{} \\ \\ \sf{ \rightarrow{0 = u + ( - gt)}{} }{} \\ \\ \sf{ \rightarrow{ - t = - \frac{u}{g} }{} }{} \\ \\ \sf{ \rightarrow{t = \frac{u}{g} }{} }{}$

When a Particle moves upwards after attaining a particular distance or height the object falls down and it travels twice in this terms,

$\sf{ \rightarrow{t = \frac{u}{g} + \frac{u}{g} }{} }{} \\ \\ \sf{{t = \frac{2(u)}{(g)} }{} }{}$

$\sf{ \purple{ \rightarrow{hence \: t = 2 \frac{u}{g} }{} }{} }{}$