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
▪️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
Answer:
********************************
▪️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
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