Question

A rock takes 2.95 s to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.3 m/s
t = 2.95 s
v = 8.3 m/s
(a):Calculate the height of the cliff in m.
(b):How long would it take to reach the ground if it is thrown straight down with the same speed?

Answer 1

fort the first one you can use the formula
\(-dy=v_isin \theta- \frac{1}{2}gt^2\)
i put "minus" because i set down as negative direction

Answer 2

ok working that all out i got 18.16

Answer 3

ok give me a sec, i'll try to solve it

Answer 4

i'm not sure but for the second one, i got the same time
O.o
i also did another solution

Answer 5

by same time you mean 2.95seconds

Answer 6

when i put 2.95 seconds in it said it was wrong

Answer 7

i told you, i'm not sure
i'm still figuring out how to calculate it

Answer 8

its no problem i appreciate all the help

Answer 9

now i get it
is it 1.25s?

Answer 10

i still used the same formula, but since it is thrown downwards
it's easier to set down as positive
\(dy=vit+ \frac{1}{2}gt^2\)
\(18.16=(8.3)t+ \frac{1}{2}(18.16)t^2\)
\(0= 4.9t^2+(8.3)t-18.16\)
where dy= 18.16
then use the quadratic formula to solve for t

Answer 11

1.25 is also wrong with this formula do you have to take the squareroot?

Answer 12

yeah, when you put in the quadratic formula, you need to take the square root of the discriminant

sorry, i guess that's all i can think of T_T
i'll ask someone to help you
@wolfe8

Answer 13

appreciate it