Question

heavy body is projected up with a velocity 19.6 metre per second find the total time needed for the same to return to its initial position​

Answer 1

Answer:

Time taken = 4 seconds

Step by step explanations :

given that,

heavy body is projected up with a velocity 19.6 m/s

here,

initial velocity of the body = 19.6 m/s

let the maximum height attained by the body be h

and the time taken by the body to reach the maximum height be t

garavitational acceleration(g) = -9.8 m/s²

now,

we have,

initial velocity(u) = 19.6 m/s

final velocity(v) = 0 m/s

[body will at rest on its maximum height]

time taken(t) = t

by the gravitational equation of motion,

v = u + gt

putting the values,

0 = 19.6 + (-9.8)t

-9.8t = -19.6

t = -19.6/-9.8

t = 2 s

now,

we know that a body takes Same time to reach the maximum height and to fall from that height

time taken = 2 × 2

= 4 seconds

so,

time taken by the body to reach it initial position

= 4 seconds

Answer 2

$\bold {\huge {Your ~answer :-}}$

$\huge {\boxed {\bold {\color {red} {4\:sec}}}}$

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Given—

Time taken = 4 seconds

Heavy body is projected up with a velocity 19.6 m/s

Now

initial velocity = 19.6 m/s

Let h me the maximum height

and where time taken t to reach max height

garavitational acceleration(g) = -9.8 m/s²

✴ g is negative as act opposite

initial velocity(u) = 19.6 m/s

final velocity(v) = 0 m/s

[body will at rest on its maximum height]

time taken(t) = t

By Law of Motion

v = u + gt

0 = 19.6 + (-9.8)t

-9.8t = -19.6

t = -19.6/-9.8

t = 2 s

As we know body always take same to reach and fall

time taken = 2 × 2 (double the time taken)

= 4 seconds

Therefore, time taken by the body to reach it initial position.

= 4 sec

$\huge{\red{\ddot{\smile}}}$

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