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When a ball thrown vertically upwards, it goes through a distance of 19.6 m. Find the initial velocity of the ball and the time taken by it to rise to the highest point? (Acceleration due to gravity, g=9.8 m/s²)?
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When a ball thrown vertically upwards, it goes through a distance of 1...
Height, s=19.6m

(i) Initial velocity, u=?

Now, when thrown vertically upwards,

Acceleration= −g=−9.8m/s2

Also, at end point, final velocity= 0m/s

⇒v2=u2+2as

So, (0)2=(u)2+2(−9.8)(19.6)

⇒u=19.6m/s

(ii) For going upward,

v=u+at

⇒0=19.6+(−9.8)(t)

⇒t=2s
Community Answer
When a ball thrown vertically upwards, it goes through a distance of 1...
Given:

Distance travelled by ball, s = 19.6 m

Acceleration due to gravity, g = 9.8 m/s²

Required:

Initial velocity of the ball, u

Time taken by the ball to reach the highest point, t

Solution:

When the ball is thrown vertically upwards, the acceleration acting on it is the acceleration due to gravity, which always acts downward.

Using the equation of motion, s = ut + 1/2 gt², where u is the initial velocity of the ball, t is the time taken by the ball and g is the acceleration due to gravity.

Finding Initial Velocity:

When the ball reaches the highest point, its final velocity becomes zero. Therefore, using the first equation of motion, v = u + gt, we can find the initial velocity of the ball.

At the highest point, v = 0

Therefore, 0 = u - gt

u = gt


Substituting the value of u in the equation of motion, s = ut + 1/2 gt², we get:

s = 1/2 gt²

t = √(2s/g)


Hence, the initial velocity of the ball is u = gt = 9.8 m/s² × √(2s/g) = √(2 × 9.8 m/s² × 19.6 m) = 19.6 m/s.

Finding Time Taken:

The time taken by the ball to reach the highest point is given by t = √(2s/g). Substituting the given values, we get:

t = √(2 × 19.6 m/9.8 m/s²) = √4 = 2 seconds.

Answer:

The initial velocity of the ball is 19.6 m/s and the time taken by the ball to rise to the highest point is 2 seconds.
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