Question

a ball is thrown upwards with a velocity 20 ms from the top of tower having a height 50 m. whai is time elasped when it passes again edge of tower​

Answer 1

Answer:

Time taken to pass edge of tower

= 2 seconds

Step by step explanations :

given that,

a ball is thrown upwards with a velocity 20 m/s from the top of tower

here,

initial velocity of the ball = 20 m/s

let the maximum height reached by ball from the initial position be h

so,

at its maximum height velocity will 0

so,

final velocity of the ball = 0 m/s

now we have,

Initial velocity(u) = 20 m/s

final velocity(v) = 0 m/s

gravitational acceleration(g) = -10 m/s²

Now,

by the gravitational equation of motion,

v = u + gt

where,

t is the time taken to reach maximum height

putting the values,

0 = 20 + 2(-10)t

-20t = -20

t = -20/-20

t = 1 s

so,

time taken to reach maximum height

= 1 seconds

now we know that,

Time taken to reach maximum height

= time taken to reach its initial position

so,

time elapsed when it passes again edge of tower = 1 + 1

= 2 seconds

Time taken to pass edge of tower

= 2 seconds

Answer 2

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

The time taken to pass edge of the tower  => 2 seconds.

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given that :

Initial velocity(u) = 20 m/s

Final velocity(v) = 0 m/s

Gravitational acceleration(g) = -10 m/s².

$\rule{300}{1.5}$

Let the maximum height reached by ball from the initial position be 'h'

We know that :

$\bigstar\; \huge \boxed{\sf u + gt = v}}$

Put the values :

$\sf =>0 = 20 + 2(-10)t\\\\=>-20t = -20\\\\=>t = \dfrac{-20}{-20}\\\\=>t =1s$

Hence,

It implies that  taken to reach maximum height  = 1 seconds.

Time elapsed when it passes again edge of tower = 1 + 1   => 2 seconds.

Time taken to pass edge of tower  = 2 seconds.