Question

A stone is thrown vertically upward with a speed of 50m/s from the edge of a cliff that is 275 m high.
1)how much later in seconds does it reach the bottom of the cliff?
2) what is the total distance in meters the stone travels before hitting the ground?

Answer 1

I have the answers if that helps, I just don't know how to get them

Answer 2

1) is 14 seconds
2) x=530

Answer 3

\(h(t)=\frac{a}{2}t^2+vt+h_0\) where
\(a\)= -9.8 m/s² or -32 ft/s², depending on the units
\(v\) = initial velocity
\(h_0\) = initial height
So your equation is
\[h(t)=-4.9t^2+50t+275\]
Set that equal to 0 and solve for t and you'll get 14.2 s.

Answer 4

Where did you get 4.9?

Answer 5

-9.8/2 = -4.9 because your length is in meters

Answer 6

I was told to use either vf=vi+at
Or
Vf^2=vi+2a × delta x

Answer 7

neither of those helps you. You're not looking for and you don't have final velocities. In your notation the formula I used would be

\[\Delta x=v_i+\frac{ 1 }{ 2 }at^2\]

Answer 8

Okaythanks

Answer 9

sorry forgot a t
\[\Delta x=v_i t+\frac{ 1 }{ 2 }at^2\]

Answer 10

How about 2?

Answer 11

You can use the second formula you put up for part 2.
The velocity is 0 at the max height, so \(v_f=0\)

\[v_f^2=v_i^2+2ad\]
\[0=(50 ~m/s)^2+2(-9.8~m/s^2)d\]
d = 127 m
so it goes 127 m on the way up.
It goes (127 m + 275 m) on the way down.
Total = 530.1 m

Answer 12

Omg,thanks so much! Now I understand a little better

Answer 13

you're welcome