A stone is dropped from the top of a building..height of building,velocity?

Question

A stone is dropped from the top of a building, 2 seconds later a second stone is thrown down at a initial velocity of 25 m/s. Both stones hit the gorund at the same time. How long does it take for the first stone to hit the ground? How high is the building? What are the speeds of each stone before they hit the ground?

Answer 1

the first stone uses a time t

so the height of buiding is H = 1/2 g t^2

the second uses t-2 the height can be written has

H = 1/2 g (t-2)^2+25 (t-2)= 1/2gt^2-2gt +2g +25t -50
taking g = 9.81 ms^-2

1/2 gt^2 = 1/2 gt^2 -19.62 t +19.62 +25t -50

6.38 t = 30.38 =4.76s time first stone

the height of buiding is H = 1/.2g (4.76)^2 =111m

speed of first stone v = gt = 9.81*4.76 =46.7 m/s

speed of second stone t-2= 4.76-2 =2.76s

v = 9.81 2.76 +25*2.76 = 90.5 m/s

Answer 2

let the time taken by first stone be t and height be h
h=1/2gt^2 and h=25(t-2) + 1/2g (t-2)^2

solve these two equations for h,t

speed of first particle=gt
speed of second=25+gt