Question

A ball is thrown up from ground with a speed of 40 m/s. Distance travelled by the particle and its velocity at the end of 5s is: JUST TELL ANSWER!!

Answer 1

velocity after 2 sec. 

v=40+(−10)t=40−10∗2=20m/s

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Answer 2

Answer:

$\bigstar{\bold{Velocity\:=\:-9\:m/s}}$

$\bigstar{\bold{Distance\:=\:77.5\:m}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Initial velocity (u) = 40 m/s
• Time taken (t) = 5s
• Acceleration(a) = Acceleration due to gravity (g) = -9.8 m/s²

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Final velocity (v)
• Distance travelled (s)

$\Large{\underline{\underline{\bf{Solution:}}}}$

Final velocity:

→ By the first equation of motion,

   v = u + gt

→ Substituting the given datas we get

  v = 40 + (-9.8) × 5

  v = -9 m/s

$\boxed{\bold{Velocity\:=-9\:m/s}}$

→ Here velocity is negative since the object is moving in the opposite direction, that is the particle is coming down.

Distance travelled:

→ By the third equation of motion,

  v² - u² = 2gs

→ Substituting the given datas,

  (-9)² - (40)² = 2 × -9.8 × s

  (81 - 1600)/-19.6 = s

  s = 77.5 m

$\boxed{\bold{Distance\:=77.5\:m}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The three equations of motion are:

• v = u + at
• s = ut + 1/2 × a × t²
• v² - u² = 2as