PLEASE HELPPP A stone is dropped from the top of a building and two seconds later another stone is thrown vertically downwards at a speed of 25 m/s. Both stones reach the ground at the same time. Find the height of the building.

Step-by-step explanation:

The equation of motion of the 1st stone which is dropped from Zero initial velocity =

d= 0.5 g t^2

The equation of motion of the 2nd stone which is dropped 1sec later with an initial velocity of 20m/s=

d= 20(t-1) +0.5 g (t-1)^2

When the second stone overtakes the 1st stone the distance covered ("d") by both the stones is the same, i.e.

0.5 g t^2 = 20(t-1) +0.5 g (t-1)^2

0.5 g t^2 = 20t-20+0.5 g (t^2+1-2t)

0.5 g t^2 = 20t-20+0.5gt^2+0.5g-gt

0= 20t-20+0.5g-gt

(g=10m/s^2)

0= 20t-20+5-10t

0=10t-15

t=1.5sec

d=0.5 x g x t^2

= 0.5 x 10 x 1.5 x 1.5

= 11.25m

Therefore, 11.25m below the top of the cliff, the 2nd stone will overtake the 1st stone.

c. (3,2,6) is the answer.