Question

1. A ball thrown up vertically returns to the throwner after 6vsec find
a)The velocity with which it was thrown up
b)The maximum height it reaches
c)its position after 4 sec

Answer 1

Answer:

The ball returns to the ground after 6 seconds.

Thus the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s

Let the velocity with which it is thrown up be u

(a). For upward motion,

v=u+at

∴ 0=u+(−10)×3

⟹u=30 m/s

(b). The maximum height reached by the ball

h=ut+

2

1

at

2

h=30×3+

2

1

(−10)×3

2

h=45 m

(c). After 3 second, it starts to fall down.

Let the distance by which it fall in 1 s be d

d=0+

2

1

at

′2

where t

′

=1 s

d=

2

1

×10×(1)

2

=5 m

∴ Its height above the ground, h

′

=45−5=40 m

Hence after 4 s, the ball is at a height of 40 m above the ground.

Answer 2

Answer:

The ball returns to the ground after 6 seconds.

Thus the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s

Let the velocity with which it is thrown up be u

(a). For upward motion,

v=u+at

∴ 0=u+(−10)×3

⟹u=30 m/s

(b). The maximum height reached by the ball

h=ut+

2

1

at

2

h=30×3+

2

1

(−10)×3

2

h=45 m

(c). After 3 second, it starts to fall down.

Let the distance by which it fall in 1 s be d

d=0+

2

1

at

′2

where t

′

=1 s

d=

2

1

×10×(1)

2

=5 m

∴ Its height above the ground, h

′

=45−5=40 m

Hence after 4 s, the ball is at a height of 40 m above the ground.

Explanation:

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