Question

A marble is thrown vertically upward with initial velocity of 40 ms–1

. Taking g = 10 ms–2

, find the maximum height

reached by the marble. What is the net displacement and the total distance covered by the marble?​

Answer 1

$\huge\mathfrak\red{ANSWER}$

IT IS GIVEN :

INITIAL VELOCITY OF MARBLE = 40m/s

ACCELERATION DUE TO GRAVITY = 10m/s^2

DISTANCE TRAVELLED = ?

DISPLACEMENT =?

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NOW FOR ANY OBJECT VERTICALLY THROWN UPWARDS AT IT MAXIMUM HEIGHT ALL NET EXTERNAL FORCES ARE EQUAL TO 0 .

THUS WE CAN SAY AT MAXIMUM HEIGHT

FINAL VELOCITY OF MARBLE = 0M/S

NOW ACCORDING TO THIRD EQUATION OF MOTION

${v}^{2} = {u}^{2} + 2ah$

HERE H IS THE MAXIMUM HEIGHT OF MARBLE.

AFTER SHIFTING SOME VALUES WE GET

$h = {v}^{2} - {u}^{2} \div 2a$

AFTER INPUTTING THE VALUES IN THIS EQUATION WE GET :

$h = 0 - {40}^{2} \div 2 \times ( - 10)$

HERE (-) IS ADDED BEFORE (g) TO REPRESENT THE NEGATIVE ACCELERATION OF MARBLE.

$h = - 1600 \div - 20$

$h = 80m$

THEREFORE THE MAXIMUM HEIGHT REACHED BY MARBLE IS 80m.

NOW SINCE DUE TO ACCELERATION DUE TO GRAVITY THE MARBLE WILL RETURN BACK TO SURFACE OF EARTH.

NOW AS INITIAL AND FINAL POSITION OF MARBLE COINCIDE SO IT'S DISPLACEMENT IS 0.

AND FINALLY AFTER REACHING THE MAXIMUM HEIGHT IT TRAVELS THE SAME DISTANCE AGAIN IN DOWNWARD DIRECTION

HENCE THE DISTANCE TRAVELLED =

$80 \times 2 = 160m$

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THUS WE GET ------------------>

MAXIMUM HEIGHT = 80m

DISTANCE TRAVELLED = 160m

DISPLACEMENT = 0m

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BY SCIVIBHANSHU

THANK YOU

STAY CURIOUS

Answer 2

Answer:

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