Question

An object is thrown vertically upwards and
rises to a height of 45 m. Calculate
[Take, g = 10 m/s2] [3]
(a) The velocity with which the object was
thrown upward.
(b) The time taken by object to reach highest
point.
(c) Total time for which object remains in air

Answer 1

Answer:

i)  We know Hmax = u²/2g ⇒ 45 = u²/20 ⇒ u² = 900 ⇒ u = 30 m/s

ii)  Time taken to rise to a height h = u/g ⇒ t = u/g ⇒ t = 30/10 s = 3 s

iii)  Total time for which the object was in the air = u/g while going up and u/g while coming down = 2u/g = 2*30/10 = 60/10 s = 6 s

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Answer 2

Given : An object is thrown vertically upwards and rises to a height of 45 m. g = 10 m/s²

To find :  The velocity with which the object was thrown upward.

(b) The time taken by object to reach highest point

(c) Total time for which object remains in air

Solution:

initial Velocity =  u  m/s

Final velocity v = 0 m/s  ( at highest point velocity become 0)

acceleration a = -g = -10 m/s²  ( as object going upward)

Distance (height)  s  = 45 m

using

V² - U² = 2aS

=> 0² - U² = 2(-10)(45)

=> U² = 900

=> U = 30 m/s

The velocity with which the object was thrown upward. = 30 m/s

V = U  + at

=> 0 = 30  + (-10)t

=> t  = 3 sec

Time  Taken

using

u = 0 here  , Starting from  max height

S = ut +  (1/2)at²

=> 45 = 0  + (1/2)10t²

=> t = 3

3 secs to come back

Total time for which object remains in air  = 3 + 3 = 6 secs

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