Solving Physics Homework: Velocity & Time of Rock in a Hole

[swiftracer]

Homework Statement

a rock is tossed straight up with a speed of 18m/s. when it returns, it fallls into a hole of 12 m deep. a) what is the velocity of the rock as it hit the bottom of the hole? b)how long is the rock in the air, from the instant it is released until it hit the bottom of the hole?

Homework Equations

a)v^2 = u^2 – 2as
b) displacement = final - initial

The Attempt at a Solution

a) and i got 12.8 and i know is wrong
b) ?

 

[Kurdt]

What have you attempted so far?

 

[kichitakashi]

solved it yet?

 

[Alcubierre]

First, know what you are looking for. In this case, it is final velocity. If you know you kinematic equations, you should see that the related equations are:

V_f² = V_i² + 2gΔy and
$Δt$ = (V_f - V_i)/g
So,

V_f² = (18 m/s)² + 2(9.81 m/s²)(12 m).

Then, take the square root and you should get your final velocity. Note, however, that it should be negative as the rock is moving toward the Earth, and that is in the negative direction.

As for the second part of the question:

$Δt$ = (V_f - V_i)/g

and that should give you the time of rock in air before it hits the bottom of the hole.
Correct me if I am wrong though, for I, myself, am unsure about this problem.

 

[PeterO]

Homework Statement

a rock is tossed straight up with a speed of 18m/s. when it returns, it fallls into a hole of 12 m deep. a) what is the velocity of the rock as it hit the bottom of the hole? b)how long is the rock in the air, from the instant it is released until it hit the bottom of the hole?

Homework Equations

a)v^2 = u^2 – 2as
b) displacement = final - initial

The Attempt at a Solution

a) and i got 12.8 and i know is wrong
b) ?

I note you listed v² = u² -2as as one of your formulas.

I don't necessarily agree with the "-" sign, I would prefer v² = u² +2as, with the last term becoming negative is a or s , but not both, were negative.
Any way: That formula is part of the set

v = u + at
v² = u² +2as
s = ut + 0.5.at²
s = vt - 0.5.at²
s = 0.5(u + v)t

That set involves the variables v, u, a, s, t with one of them missing from each equation in turn.

For this problem, you just write down the variables you know, add on the one you want to find out, then use the formula involving those variables.

For this rock: let up be positive:

(a) u = 18, a = -9.8, s = -12 and we want v

we need the second formula, which involves v, u, a & s

v² = 18² + 2 x -9.8 x -12

Solve that and you will have your answer:

(b) u = 18, a = -9.8, s = -12 and we want t

That means use the 3rd formula

-12 = 18t - 4.9t²

Unfortunately a quadratic equation to solve; but if you are confident of your answer to part (a), you could add that v value into the mix and use one of the equations not involving quadratics, like the first one.

 

[micromass]

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