Question

A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60m above the ground. How much time elapses between the instant of release and the instant of impact with the ground?

Answer 1

i did \[\implies 60 = 20t^2 + 4.8t\]

Answer 2

then my calulator said t = 1.6

but the right answer was t - 2

Answer 3

t = 2*

Answer 4

can anyone explain where i went wrong?

Answer 5

9.8/2=4.9

Answer 6

uhh oops

Answer 7

still doesnt give right answer

Answer 8

but i got it 2

Answer 9

how?

Answer 10

which equation u used?

Answer 11

\[t=\frac{ -20 \pm \sqrt{(20)^2+4*4.9*60} }{ 9.8 }=2.01\]

Answer 12

\[x = V_1 t^2 + \frac{at}2\]

Answer 13

and the equation is \[4.9t^2+20t-60=0\]

Answer 14

i think i got it backwards...

Answer 15

hmm no wonder

Answer 16

i did have it backwards

Answer 17

u took a,b,c values incorrect @ghazi

Answer 18

@hartnn it is correct

Answer 19

use s= ut + 1/2 gt^2...

Answer 20

okk, i thought lg's equation was correct.

Answer 21

i interchanged my signs incorrectly..that's why i cant get 2