Physics, asked by kdev6363, 5 months ago

Whenever a body is thrown up with a certain velocity, the upward motion is opposed by the

gravitational pull of earth and also by the resistance of air. Therefore, velocity of the body goes

on decreasing. When this velocity becomes zero, the body cannot rise further. It has attained,

what is called ‘maximum height’. From this height, the body begins to fall downwards under the

action of gravity.

Assume that the air resistance is zero or negligible.What is the acceleration of the body at the highest point?​

(a) +9.8 m/s2 in the downward direction

(b) -9.8 m/s2

in the downward direction

(c) +9.8 m/s2

in the upward direction

(d) -9.8 m/s2

in the upward direction​

Answers

Answered by satyamkumar5428
19

(d) –9.8 m/s² in the upward direction

Explanation:

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Answered by ExᴏᴛɪᴄExᴘʟᴏʀᴇƦ
35

Answer :

\quad ● +9.8 m/ in the downward direction [Option A]

━━━━━━━━━━━━━━

  • When the body is thrown up we know that the body deceletes due to the gravitational force and as it deceletes it would be -9.8 m/s²

  • When the body is at the maximum height with a velocity of 0 m/s then the gravitational force would pull the ball towards the earth which means it's accelerating and as it is accelerating, i.e the velocity of the ball is increasing

  • As the Ball's velocity is increasing the acceleration would be positive that is +9.8 m/s²

  • And as the ball comes towards the earth, that is in the downward direction the answer to the question becomes +9.8 m/s² in the downward direction!!

\setlength{\unitlength}{1mm}\begin{picture}(8,2)\thicklines\multiput(9,1.5)(1,0){50}{\line(1,2){2}}\multiput(33,7)(0,4){12}{\line(0,1){2}}\multiput(36,10)(0,4){12}{\line(0,1){2}}\put(10.5,6){\line(3,0){50}}\put(34,60){\circle*{10}}\put(2,15){\large\sf{Thrown up = -ve}}\put(37,55){\large\sf{Coming down = +ve}}\put(22,61){\large\textsf{\textbf{Ball}}}\put(36,12){\vector(0, - 4){5}}\put(33,50){\vector(0,4){5}}\end{picture}


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