Sociology, asked by rockstarxx, 7 months ago

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-2

Answers

Answered by Anonymous

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▪️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.

▪️Given:

▪️Initial velovity of ball 'u' = 19.6 m/s

▪️Time taken by ball to reach ground 't' = 5sec.

▪️ Gravity 'g' = 9.8 m/s

Solution:

▪️we know that,

▪️⇒ g = \frac{v}{t}

▪️ ⇒ t = \frac{v}{g}

= \frac{19.6}{9.8}

▪️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 + \frac{1}{2} a t^{2}

▪️⇒ s = 19.6 x 1 + \frac{1}{2} x 9.8 x 1^{2}

▪️⇒ 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

Answered by jiya9614

Answer:

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