Question

A ball is thrown vertically upwards from the top
of a tower with an initial velocity of 19.6 ms.
The ball reaches the ground after 5 s. Calculate :
(i) the height of the tower, (ii) the velocity of ball
on reaching the ground. Take g = 9.8 m s.​

Answer 1

Answer:

total time taken is 5 seconds so time to fall freely from the max height..

2.5 seconds...

so using S = 1/2at^2 = 1/2× 10× (2.5)^2=

1)31.25m

using ,v= u+at, u= 0 in free fall ....

v= at= 10×2.5 = 25 m/s

Answer 2

Answer:

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Explanation:

here,

From the top of the tower when the ball is thrown and when it reaches the bottom again,

the Time taken total = 2u/g => 4 seconds

that means it takes 5-4 =1 s to travel the tower height.

and max. height it takes = > u^2/2g => 19.6*19.6/2*9.8 => 19.6 meter.

From first law of eqn. we get,

v = 19.6 – g*4 => -19.6 m/s.

Hence from second’s law of eqn. we get,

h = 19.6*1 + 0.5*9.8*1^1 = > 24.5 meter.

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