Physics, asked by sangithakailash7451, 10 months ago

From top of a building 125 m high a ball is thrown horizontally with speed 10 m/s. the time taken by it to reach the ground is.(take g=10 m/s² ) *

Answers

Answered by Anonymous
73

Given :

▪ Height of building = 125m

▪ Initial horizontal velocity = 10mps

▪ Acc. due to gravity = 10m/s²

To Find :

▪ Time taken by ball to reach at the ground.

Solution :

☞ This question is completely based on the concept of Height to Ground projectile motion.

☞ In this type of motion, time taken by object to reach at the ground depends only on vertical motion.

☞ Initial velocity for vertical motion = zero

☞ Horizontal component of velocity remains constant throughout the motion but vertical component of velocity doesn't remain constant as gravitational acceleration acts in the downward direction continuously.

\dashrightarrow\sf\:H=u_yt+\dfrac{1}{2}a_yt^2\\ \\ \dashrightarrow\sf\:H=(0\times t)+\dfrac{1}{2}gt^2\\ \\ \dashrightarrow\sf\:H=\dfrac{1}{2}gt^2\\ \\ \dashrightarrow\sf\:t=\sqrt{\dfrac{2H}{g}}\\ \\ \dashrightarrow\sf\:t=\sqrt{\dfrac{2\times 125}{10}}\\ \\ \dashrightarrow\sf\:t=\sqrt{25}\\ \\ \dashrightarrow\underline{\boxed{\bf{\blue{t=5s}}}}\:\orange{\bigstar}

Answered by Anonymous
42

Given:-

\sf\ Height\: of\: a \:building\:= 125m

\sf\ Speed = 10m/s

\sf\ According\: due\: to\: gravity\: = 10m/s^2

To find:-

\sf\ Time\: taken\: by\: ball \:to\: reach \:the \:ground.

Required Solution:-

\sf\ First\: we \:have\: to\: final\: final\: velocity

\sf\ Using\: the\: third\: equation \:of\: motion:-

\dashrightarrow\: \sf\ v^2 = u^2+ 2GH

\dashrightarrow\: \sf\ v^2 = 0^2 + 2×10×125

\dashrightarrow\: \sf\ v^2 = 2500

\dashrightarrow\: \sf\ v = 50m/s

\sf\ Then,\: we \:have\: to\: find\: Time.

\sf\ Using\: the\: first\: equation= v=u+at

\dashrightarrow\: \sf\ t = v - u / G

\dashrightarrow\: \sf\ t = 50 - 0 / 10

\dashrightarrow\: \underline{\boxed{\bf{\red{t=5s}}}}

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