physics - again, but last one tonight

[tjokipii]

Thanks so much for the help on my other problem. This is the last one that I can't get this evening. Again, thanks.

When a kid drops a rock off the edge of a cliff, it takes 4.0 s to reach the ground below. When he throws the rock down, it srikes the ground in 3.0 s. What initial speed did he give the rock?

Thanks -

Terry[/i]

 

[Guest]

kid drops a rock off the edge of a cliff, it takes 4.0 s to reach the ground below

Using the terms from the last posting

u= 0 , a=9.8 , t= 4 s = ?

s= ut + (1/2) a t^2

s = 0 + (1/2) 9.8 (4)^2

s= 78.4 m


Now for the throw case the initial velocoity is unknow

u = ? , s =78.4 , a=9.8 t=3

s= ut + (1/2) a t^2

(s- (1/2) a t^2 )/ t = u

put the numbers in and solve

 

[stapel]

Use the basic vertical projectile equation, s(t) = -(1/2)gt² + v₀t + h₀, and, using the gravity information (the value of "g" will depend on what units you're using, and that information does not appear in your post) and the time, solve for the initial height from which the rock was dropped.

Use this height h₀ and the gravity information, along with the provided time information, to solve for the initial velocity v₀ ("u", in the other thread).

Eliz.

 

[tjokipii]

to everyone who helped

I'm leaving to go work with the formulas you gave me. This site is so great. I'm so glad I found it. Our physics class doesn't have a text book and he hasn't given us any formulas. I don't know what I would do without the internet. Thanks everyone. I'll talk to you soon.

Terry

 

[Guest]

The 3 equations of motion that you need to use are these

v^2 = u^2 + 2 a s

v = u + a t

s= ut + (1/2) a (t^2)

where
u= initial velocity (m/s)
v = final velocity (m/s)
t = time (s)
s = displacement (m)
a = acceleration (m/s^2)

Hint -- if you have any 3 of the above variables the problem is solved by just using the correct equation.