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

AnswEr :

⠀

Time Elapsed is 4 Second when it Passes again Edge of Tower.

⠀

Explanation :

• Initial Velocity (u) = 20 m/s

• Final Velocity (v) = 0 m/s

• Gravitational Acceleration (g) = - 10 m/s²

⋆ By First Equation of Gravitational Motion

➟ v = u + gt

➟ 0 = 20 + (- 10) × t

➟ 0 = 20 - 10t

➟ 10t = 20

➟ $\bold{ t = \cancel{\dfrac{20}{10}}}$

➟ t = 2 second

⠀

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⠀

• It Takes 2 Second to Go Upward , So it will must take 2 Second to Go Downward. The Total Time Will Be :

➡ Total Time Taken = 2 + 2

➡ Total Time Taken = 4 second

⠀

$\large\therefore$ Time Elapsed is 4 Second when it Passes again Edge of Tower.