Question

A ball is thrown vertically upwards from a point on the
ground. It was observed at a height h twice with a time
interval At. The initial velocity of the ball is​

Answer 1

Answer

u

u 1

u 1

u 1 =

u 1 = 2

u 1 = 21

u 1 = 21

u 1 = 21

u 1 = 21 8gh+g

u 1 = 21 8gh+g 2

u 1 = 21 8gh+g 2 (△t)

u 1 = 21 8gh+g 2 (△t) 2

Explanation:

If the ball is reaching height h then, displacement=h

Time taken=t

1

... (say)

Acceleration=−g

Initial velocity=u

1

∴from the equations of motion

S=ut+

2

1

at

2

∴h=u

1

t

1

+

2

1

(−g)t

1

2

.....(i)

If the ball is reaching height 2hthen, displacement=2h

Time taken=t

1

+△t

Acceleration=−g

Initial velocity=u

2

Therefore from equations of motion

2h=u

2

(t

1

+△t)−

2

1

g(t

1

+△t)

2

.....(ii)

Also we know from the equations of motion that

v=u+at∴0=u

1

−gt

1

....(iii)

& 0=u

2

−g(t

1

+△t).....(iv)

∴u

1

=gt

1

,u

2

=gt

1

+g△t=u

1

+g△t

∴From equations (i),(ii),(iii),(iv)

we get(ii)−(i)=h={(u

1

+g△t)(t

1

+△t)−

2

1

g(t

1

+△t)

2

−u

1

t

1

−

2

1

gt

1

2

}

∴u

1

=

2

1

8gh+g

2

(△t)

2

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Answer 2

Answer:

Aakash AIATS de rhe ho na

Explanation:

answer is option (1)