Question

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Sub :- Physics

Class :- 9th

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Q. A ball is thrown upward with some initial speed it goes up to a height of 19.6m and then returns. Find :

(a) the initial speed

(b) the time taken in reaching the highest point

(c) the velocity of the ball one second before and one second after it reaches the maximum height

(d) the time taken by the ball to return to its original position.


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Answer 1

$\huge{\mathcal{\star\:Hey\:Mate\:\star}}$

$<b><i>Solution:$

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(a) At the maximum hight, the velocity becomes zero. Taking the upward direction as positive,

v² = u² - 2gs

=> 0² = u² - 2 × (9.8 m/s²) × (19.6m)

=> u² = (2 × 9.8 × 19.6) m²/s²

=> u = 19.6 m/s

(b) We have,

v = u - gt

=> 0 = (19.6 m/s) - (9.8 m/s² ) × t

=> t = 19.6/9.8

=> t = 2 sec

So, the particle takes 2 s to reach the highest point.

(c) The ball reaches the maximum height at t = 2s. We have to find the velocity at t = 1 s and t = 3 s.

We have,

v = u - gt

At t = 1 s,

0 = (19.6 m/s) - (9.8 m/s²) × (1s)

=> 19.6 - 9.8

=> 9.8 m/s

As v is positive, the particle is going up with the speed of 9.8 m/s.

At t = 3s,

v = u - gt

=> v = (19.6 - 9.8) × 3s

=> v = 19.6 - 29.4

=> v = - 9.8 m/s.

As v is negative, the particle is coming down with a speed of 9.8 m/s.

(d) When the ball returns to its original position, its displacement s = 0.

Using,

s = ut - 1/2 gt²

0 = (19.6)t - $\frac{1}{2}$ × ( 9.8)t²

As t ≠ 0

19.6 = (4.9 m/s² )t

t = $\bold{\frac {19.6}{4.9}}$

$\therefore$ t = 4 sec.

So, the ball takes 4s to return to its original position.

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Answer 2

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refer to the attachment

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hope helped

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