Question

A Ball was thrown up Vertically returns to the Thrower after 6 Seconds. Find -

(a) The Velocity with which it was Thrown up.
(b) The maximum Height it reaches, and
(c) It's Position after 4 Seconds.

(Acceleration due to Gravity, g = 9.8 m/s²)​

Answer 1

Explanation:

(A) SINCE VALUE OF g REMAINS SAME THEREFORE THE MAXIMUM TIME BALL TAKES TO REACH THE FLOOR

= TOTAL TIME / 2

$= 6 \div 2$

= 3 SECONDS.

SINCE THE BALK COMES BACK TO THW THROWER THIS MEANS THAT THE FINAL Velocity BECOMES ZERO (0).

NOW SOME KEYWORDS ABOUT BALL=>>

FINAL VELOCITY = v = 0m/s

INITIAL VELOCITY = u = ?

ACCELERATION = g = 9.8m/s

TIME TAKE TO ACCELERATE = 3 SECS.

NOW ACCORDING TO FIRST EQUATION OF MOTION :

v = u + at

-u = at + v

u = -(at + v)

AFTER Inputting THE UPPER VALUES IN EQUATION WE GET :

u = -(gt + v)

u = - ( -9.8(3) +0)(-) before (g) indicates retardation.

u = 9.8 × 3

u=29.4 m/s

THEREFORE THE VELOCITY WITH WHICH BALL WAS THROWN IS 29.4m/s

______________________________________

(B) MAXIMUM HEIGHT IT REACHES CAN BE SAID DISTANCE IT COVERS IN UPWARD DIRECTION - IT WILL BE (s).

NOW ACCORDING TO THIRD EQUATION OF MOTION.

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

THEREFORE ON SHIFTING (u) and (2a) ON L.H.S WE GET:

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

NOW, AFTER INPUTTING THE VALUES IN THIS EQUATION WE GET:

${0}^{2} - {29.4}^{2} \div 2(9.8) = s$

$s = 864.36 \div 19.6$

$s = 44.1m$

THEREFORE THE MAXIMUM HEIGHT COVERED BY BALL IS 44.1M.

______________________________________

(C) PAY ATTENTION HERE:

IF THE BALL REACHES MAXIMUM HEIGHT IN UPWARD DIRECTION IN 3RD SEC.

THIS MEANS ON 4th SECOND IT WILL HEAD TO DOWNWARD DIRECTION.

NOW DISTANCE COVERED BY BALL IN 1s

$s = 1 \div 2 + a {t}^{2}$

$s = 9.8 \times 1 \times 1 \div 2$

$s = 4.9m$

NOW ITS POSITION OF BALL IN 4SEC

$= 44.1 - 4.9$

THATS BECAUSE IT'S POSITION IS MAX. AT 3RD SEC AND AT 4 IT WILL BE 3-1.

THEREFORE THE POSITION OF BALL AT 4TH SEC WILL BE 39.2 m.

______________________________________

BY SCIVIBHANSHU

THANK YOU

STAY CURIOUS

Answer 2

Answer:

(A) SINCE VALUE OF g REMAINS SAME THEREFORE THE MAXIMUM TIME BALL TAKES TO REACH THE FLOOR

= TOTAL TIME / 2

= 6 \div 2=6÷2

= 3 SECONDS.

SINCE THE BALK COMES BACK TO THW THROWER THIS MEANS THAT THE FINAL Velocity BECOMES ZERO (0).

NOW SOME KEYWORDS ABOUT BALL=>>

FINAL VELOCITY = v = 0m/s

INITIAL VELOCITY = u = ?

ACCELERATION = g = 9.8m/s

TIME TAKE TO ACCELERATE = 3 SECS.

NOW ACCORDING TO FIRST EQUATION OF MOTION :

v = u + at

-u = at + v

u = -(at + v)

AFTER Inputting THE UPPER VALUES IN EQUATION WE GET :

u = -(gt + v)

u = - ( -9.8(3) +0)(-) before (g) indicates retardation.

u = 9.8 × 3

u=29.4 m/s

THEREFORE THE VELOCITY WITH WHICH BALL WAS THROWN IS 29.4m/s

______________________________________

(B) MAXIMUM HEIGHT IT REACHES CAN BE SAID DISTANCE IT COVERS IN UPWARD DIRECTION - IT WILL BE (s).

NOW ACCORDING TO THIRD EQUATION OF MOTION.

{v}^{2} = {u}^{2} + 2asv

2

=u

2

+2as

THEREFORE ON SHIFTING (u) and (2a) ON L.H.S WE GET:

{v}^{2} - {u}^{2} \div 2a = sv

2

−u

2

÷2a=s

NOW, AFTER INPUTTING THE VALUES IN THIS EQUATION WE GET:

{0}^{2} - {29.4}^{2} \div 2(9.8) = s0

2

−29.4

2

÷2(9.8)=s

s = 864.36 \div 19.6s=864.36÷19.6

s = 44.1ms=44.1m

THEREFORE THE MAXIMUM HEIGHT COVERED BY BALL IS 44.1M.

______________________________________

(C) PAY ATTENTION HERE:

IF THE BALL REACHES MAXIMUM HEIGHT IN UPWARD DIRECTION IN 3RD SEC.

THIS MEANS ON 4th SECOND IT WILL HEAD TO DOWNWARD DIRECTION.

NOW DISTANCE COVERED BY BALL IN 1s

s = 1 \div 2 + a {t}^{2}s=1÷2+at

2

s = 9.8 \times 1 \times 1 \div 2s=9.8×1×1÷2

s = 4.9ms=4.9m

NOW ITS POSITION OF BALL IN 4SEC

= 44.1 - 4.9=44.1−4.9

THATS BECAUSE IT'S POSITION IS MAX. AT 3RD SEC AND AT 4 IT WILL BE 3-1.

THEREFORE THE POSITION OF BALL AT 4TH SEC WILL BE 39.2 m.

______________________________________