A ball is thrown vertically upwards from the top of a
tower with an initial velocity of 19.6 m s-1. 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-2
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I not understood from there
Answers
Explanation:
1. The height of the tower is 24.5 m.
2. The velocity of the ball on reaching the ground is 29.4 m/s.
Explanation:
Given:
Initial velovity of ball 'u' = 19.6 m/s
Time taken by ball to reach ground 't' = 5 sec.
Gravity 'g' = 9.8 m/s
Solution:
we know that,
⇒ g = 9.8
⇒ t =
=
So, ⇒ t = 2 sec.
it means time taken to reach max. height is 2 seconds.
It means 2 seconds are taken by the ball to reach back to tower and rest of 1 second to reach to ground.
it means first two seconds are taken to make final velocity 'v' is zero (max. height).
⇒ v = 0 m/s
Time taken is 1 second to calculate height of tower (written above).
So, using equation of motion,
⇒ s=ut +1/2 at²
⇒ s = 19.6 x 1 + x 9.8 x
⇒ s = 24.5 m
From zero initial velocity at time 2 seconds to ground where time is 5 seconds. Now, time taken is 3 seconds from max. height to ground.
now, ⇒ v = u + at
⇒ v = 0 + 9.8 x 3
⇒ v = 29.4 m/s
Explanation:
height of the tower = the 24.5m
v of the ball = 29.4