Question

A ball is thrown vertically upwards 40 metres per second : calculate the greatest height attained and its velocity after 3 seconds

Answer 1

At the highest point, velocity of the ball is 0.

Using the equations of motion:

v² = u² + 2aS

⇒ 0 = (40)² + 2(-g)S

⇒ -1600 = 2(-g)S

⇒ 800/g = S

     If |g| = 10,   S = 800/10 = 80 m

     if |g| = 9.8,  S = 800/9.8 = 81.63 m

Using v = u + at   for velocity at t = 3

   v = (40) + (-10)(3)  or  (40) + (-9.8)(3)

      = 40 - 30     or   40 - 29.4

      = 10 m/s      or   10.6 m/s

Answer 2

Answer:

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Given :-

• A ball is thrown vertically upwards 40 meters per sec.

To find :-

• calculating greatest height attained.

Explanation :-

• Here by using (equations of motion formula we get that)
• ${v}^{2} = {u}^{2} + 2as$
• Here , one thing we want to know that, at the highest height velocity of ball will be zero .
• Therefore,

• Applying the values we get that,

• $0 = {40}^{2} + 2(g)s$
• $s = \frac{800}{g}$
• If the value of g = 10,
• $s = \frac{800}{10 } = 80m$
• So, by the first equation of motion for velocity at 3 formula we get that,

• $v = u + at$
• Now applying the values we get,

• $v = 40 + ( - 10)(3)$
• $v = 40 - 30$
• $v = 10m \: per \: sec$

Approximately we get the value as, 10 m/s or 10.6m/s in another method as also we can find ok.

Used formula :-

• First equations of motion
• and we can take the value of g as approximately 10 or 9.8m/s.
• So finally we get the greatest height attained by the ball

Hope it helps u mate .

Thank you .