Math, asked by Sakshi5658, 10 months ago

A ball is gently dropped from a height of 20m if its velocity increases uniformly at the rate of 10 metre per second square with what velocity will it strike the ground? After what time will it strike the ground?

Answers

Answered by Anonymous
129

  \huge{ \mathtt{GivEn:-}}

  • Initial velocity, (u) = 0
  • Final velocity, (v) = ?
  • Acceleration, (a) = 10 m/s²
  • Distance, (s) = 20m

 \large{ \mathtt{NeEd \:  tO  \: fiNd:-}}

  • Velocity at which the ball strike the ground.
  • Time it will take to strike the ground.

 \large \mathtt{forMula \:  usEd:-}

  • v² = u² + 2as
  • v = u + at

 \huge{ \mathtt{ \underline{ \red{ \: ExplaNatiOn:-}}}}

 \small{ \mathtt{ \bold{CalculaTinG  \: veloCity  \: oF  \: thE \:  ball}}}

As we know that;

 \large \boxed{ \purple{ \mathtt{ {v}^{2}  =  {u}^{2}  + 2as}}}

Putting values in this formula,

We get;

v² = u² + 2as

v² = (0)²+2×10×20

v² = 0+400

v² = 400

v = √400

v = 20 m/s

 \huge{ \mathrm{ \blue{NOW,}}}

 \small{ \mathtt{ \bold{CalculaTinG  \: tiMe  \: takeN;}}}

As we know that;

 \large{ \boxed{ \pink{v = u + at}}}

Putting values in this formula,

We get;

v = u + at

20 = 0+10×t

10t = 20

t = 20/10

t = 2 sec

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