Question

a billard ball is dropped from the roof of a 1300' tall building. How far has the ball fallen after 7 sec. and How long will it take for the ball to hit the ground (round to nearest sec.)

Answer 1

not accounting for reality; -16t^2+1300 at t=7

Answer 2

s=ut+at^2/2
s=9.81*49/2
1300=9.81t^2/2
now the 2nd equation gives u the distance and 3rd gives u time....

Answer 3

i thought the equation was s=16t^2+Vot+So or is that the same thing u just did?

Answer 4

well; -16t^2 since gravity is pulling against it

Answer 5

at 16t^2; it never falls down; only up

Answer 6

oh! see conventions vary...if u go with amistre method then s is -ve wen coming down and +ve wen goin up...

Answer 7

4.6t^2 is the metric right?

Answer 8

yes! :)

Answer 9

Distance fallen after t secs = S = ut + (1/2)gt^2 - (1)
where u = 0 since starts from rest, g = 32ft/sec^2 and t = 7 secs .
Hence the distance in 7 secs = 0x7 + (0.5) x 32x7x7
0 + 16 x 49 = 784 ft

It will hit the ground when it has travelled the distance S=1300 ft
Substiting the same eqn :(1) 1300 = 0x7 + (0.5)x32xt^2

ie 1300 = 0 + 16 t^2

1e 1300 = 16 t^2

or t^2 = 1300/16 Hence t = sq rt (1300) / sq rt 16

= 36.0555/4 = 9.01 secs

= 9 secs (nearest secs)

Answer 10

Substituting in the same equation