Question

16. A ball P is dropped from the top of a tall tower and falls freely towards the earth.
Another ball Q is dropped from the top of the same tower exactly two seconds later.
Then the separation between P and Q two seconds after the dropping of the second ball
Q is (take g = 10 m/s2)
1) 10 m
2) 40 m
3) 60 m
4) 80 m​

Answer 1

Answer:

distance travelled  by ball in time t  is 21gt2

separation : s=21g42−21g32=5×16−5×7=35

Answer 2

Given:

A ball P is dropped from the top of a tall tower and falls freely towards the earth.

Another ball Q is dropped from the top of the same tower exactly two seconds later.

To find:

Separation between P and Q after 2 seconds of dropping Q ball.

Calculation:

So , time for which P ball has fallen = 2+2 = 4 sec.

Let Displacement be d_(P):

$\therefore \: d_{P} = \dfrac{1}{2}g{(t_{P})}^{2}$

$= > \: d_{P} = \dfrac{1}{2} \times 10 \times {4}^{2}$

$= > \: d_{P} = \dfrac{1}{2} \times 10 \times 16$

$= > \: d_{P} = \dfrac{160}{2}$

$= > \: d_{P} = 80 \: m$

Time for which ball Q falls is 2 secs.

Let displacement be d_(Q):

$\therefore \: d_{Q} = \dfrac{1}{2} g {(t_{Q})}^{2}$

$= > \: d_{Q} = \dfrac{1}{2} \times 10 \times {(2)}^{2}$

$= > \: d_{Q} = \dfrac{1}{2} \times 10 \times 4$

$= > \: d_{Q} = \dfrac{40}{2}$

$= > \: d_{Q} = 20 \: m$

So , separation between P and Q :

$\therefore \: d = d_{P} - d_{Q}$

$= > \: d =80 - 20$

$= > \: d =60 \: m$

So, separation is 60 m.

HOPE IT HELPS.