Question

A ball starts from rest on a horizontal floor and is given an acceleration of 2m/s^2. It hits the wall at a distance d from it. In the last second before it hits the wall, it covers a distance of 15m. Find the distance d​

Answer 1

Answer:

When the stone is dropped from the balloon, its initial velocity is the same as the velocity of the balloon at that instant.

The upward acceleration of the balloon is a=2m/s

2

The balloon starts from rest, so u= 0 m/s

The balloon rises for 1 s before the stone is dropped, hence we have

v =u + at

v=0+=2 m/s in the vertically upw ard direction.

This is the initial velocity of the stone after being dropped.

The distance moved up by the balloon in 1 second is

v

2

=u

2

+2as

v

2

=2as

∴S=

2a

v

2

=

4

4

=1m

Hence, the stone falls by 1 m before hitting the ground

Now, the acceleration on the stone after being dropped is g=9.8m/s

2

Hence, for the stone,

we have

u=−2m/s;s=1m;a=9.8m/s

2

;t=1s

u is negative as it is in the upward direction

Hence, using third equation of motion,

we get time as

s=ut+

2

1

at

2

1=−2t+

2

1

×9.8t

2

4.9t

2

−2t−1=0

∴=t=

2×9.8

2

−

+

(−2)

2

−4×4.9×(−1)

=

2×9.8

2

−

+

4+19.6

=

9.8

2

−

+

23.6

Since, time cannot be negative, we have

t=

9.8

2+

23.6

=0.7s

Hence, the stone will strike the ground 0.7 s after being dropped from the balloon.

hence (B) option is correct

Answer 2

Answer:When the stone is dropped from the balloon, its initial velocity is the same as the velocity of the balloon at that instant.

The upward acceleration of the balloon is a=2m/s  

2

 

The balloon starts from rest, so u= 0 m/s

The balloon rises for 1 s before the stone is dropped, hence we have

v =u + at

v=0+=2 m/s in the vertically upw ard direction.

This is the initial velocity of the stone after being dropped.

The distance moved up by the balloon in 1 second is

v  

2

=u  

2

+2as

v  

2

=2as

∴S=  

2a

v  

2

 

​  

=  

4

4

​  

=1m

Hence, the stone falls by 1 m before hitting the ground

Now, the acceleration on the stone after being dropped is g=9.8m/s  

2

 

Hence, for the stone,  

we have  

u=−2m/s;s=1m;a=9.8m/s  

2

;t=1s

u is negative as it is in the upward direction

Hence, using third equation of motion,

we get time as

s=ut+  

2

1

​  

at  

2

 

1=−2t+  

2

1

​  

×9.8t  

2

 

4.9t  

2

−2t−1=0

∴=t=  

2×9.8

2  

−

+

​  

 

(−2)  

2

−4×4.9×(−1)

​  

 

​  

=  

2×9.8

2  

−

+

​  

 

4+19.6

​  

 

​  

=  

9.8

2  

−

+

​  

 

23.6

​  

 

​  

 

Since, time cannot be negative, we have  

t=  

9.8

2+  

23.6

​  

 

​  

=0.7s

Explanation: