Physics, asked by sanyamsaini3886, 2 months ago

A ball is gently dropped from a height of 30m. If its velocity increase uniform at the rate of 10m/s with what veloscity will it stricks the ground ? After what time will it stricke the ground

Answers

Answered by RISH4BH
69

\red{\bigstar}\underline{\underline{\textsf{\textbf{ Given :- }}}}

  • A ball is gently dropped from a height of 30m.
  • Its velocity increase at the rate of 10m/.

\red{\bigstar}\underline{\underline{\textsf{\textbf{ To Find :- }}}}

  • The velocity with it strikes the ground .
  • The time after which it strikes the ground .

\red{\bigstar}\underline{\underline{\textsf{\textbf{ Solution :- }}}}

Here the ball is dropped from the height of 30m . Hence it will have a initial Velocity of 0m/s . The distance travelled by the ball is equal to the height of the tower and that is 30 metre . Now using the Third equation of motion , we have ,

\sf\dashrightarrow 2as = v^2-u^2\\\\\\\sf\dashrightarrow 2 * 10m/s^2 * 30m = v^2 - (0m/s)^2 \\\\\\\sf\dashrightarrow v^2 = 600 m^2/s^2\\\\\\\sf\dashrightarrow v =\sqrt{600m^2s^{-2}}  \\\\\\\sf\dashrightarrow \underset{\blue{\sf Required\ Velocity }}{\underbrace{\boxed{\pink{\frak{ Final\ Velocity \ (v) = 24.49 \ m/s \approx 24.5 m/s}}}}}

\qquad\rule{200}2

Now we need to find the time of fall . Here the acceleration of the ball is 10m/ . Initial Velocity is 0m/s and the final velocity is 24.5 m/s . Now using the First equation of motion ,

\sf\dashrightarrow v = u + at \\\\\\\sf\dashrightarrow t =\dfrac{ v - u }{a} \\\\\\\sf\dashrightarrow t  =\dfrac{ 24.5 m/s }{10m/s^2} \\\\\\\sf\dashrightarrow\underset{\blue{\sf Required\ time }}{\underbrace{\boxed{\pink{\frak{ Time (t) = 2.45 \ s }}}}}

\qquad\rule{200}2

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