# Brain Fart :/

• March 23rd 2006, 10:09 AM
tnkfub
Brain Fart :/
a stone is tossed doown with a speed of 8 m/ s from the edge of a cliff 63 m high. how long will it take to hit the foot of the cliff?
a= -9.8 m/s ^2
vi= -8 m/s
s= 63m

63m= - 8 m/s(t) + 1/2(9.8m/s^2)(t)^2

does this look right ?

and if so how do i solve for t?

Just confused can someone help me :/ :confused:
• March 23rd 2006, 11:39 AM
earboth
Quote:

Originally Posted by tnkfub
a stone is tossed doown with a speed of 8 m/ s from the edge of a cliff 63 m high. how long will it take to hit the foot of the cliff?
a= -9.8 m/s ^2
vi= -8 m/s
s= 63m

63m= - 8 m/s(t) + 1/2(9.8m/s^2)(t)^2

does this look right ?
and if so how do i solve for t?
Just confused can someone help me :/ :confused:

Hello,

the stone starts at a height of 63 m. Then it is loosing height by being tossed and by falling, until it hits the ground, that means the height is zero(0).

$0= 63- 8\frac{m}{s} \cdot t- \frac{1}{2} \cdot 9.81 \frac{m}{s^2} \cdot t^2$

This is a quadratic equation in t. I suppose that you know how to solve a quadratic equation.

You'll get $x_1 \approx 2.86 s\ \vee \ x_2=-4.49 s$

The negative solution isn't very realistic.

In comparison: If the stone were only fallen down, it had taken 3.58 s to come down. So the tossing gave the extra kick.

Greetings

EB
• March 23rd 2006, 03:13 PM
topsquark
Quote:

Originally Posted by tnkfub
a stone is tossed doown with a speed of 8 m/ s from the edge of a cliff 63 m high. how long will it take to hit the foot of the cliff?
a= -9.8 m/s ^2
vi= -8 m/s
s= 63m

63m= - 8 m/s(t) + 1/2(9.8m/s^2)(t)^2

does this look right ?

and if so how do i solve for t?

Just confused can someone help me :/ :confused:

Earboth didn't leave much to comment on, but I want to make a suggestion, considering the formula you came up with. It might seem like over-kill, but it's good practice to always draw a diagram and select a positive direction.