Question

A stone is thrown vertically upward with a velocity of 20m/s. If there is no resistance of air then calculate
(a) the maximum height reached by stone,
(b) total time taken by stone to return to its original position
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Answer 1

Answer:

According to the equation of the motion under gravity

v

2

−u

2

=2gs

u=initial velocity of the stone=40m/s

v= Final velocity of the stone=0m/s

Let h be the maximum height attained by the stone

Therefore,

0

2

−40

2

=2(−10)h

h=(40×40)/20=80

Therefore, total distance covered by the stone during its upward and downward journey=80+80=160m

Net displacement during its upward and downward journey=80+(−80)=0

Explanation:

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Answer 2

Answer:

u=20m/s^2

g/a=-9.8/10m/s^2

h=?

v2-u2 =2ax

since it's a free fall object

v2-u2 =2gh

(at hmax v=0)

0^2 -(40)^2=2*(-10)*h

-1600=-20h

h=80m

(b). v=u+at

0=20-10*t

10t=20

t=2s

Explanation:

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