Finding Time Using Displacement and Acceleration

[Paincake]

Homework Statement
"A body is thrown downward with an initial speed of 20 m/s on Earth. What is the time required to fall 300 m?"

I tried finding kinetic and gravitational energy, but then I realized I don't have any mass to use it with (1/2)*m*v^2 and m*g*h

How can I solve this problem without guessing and checking?

 

[collinsmark]

Hello Paincake,

Welcome to physics forums!

Homework Statement
"A body is thrown downward with an initial speed of 20 m/s on Earth. What is the time required to fall 300 m?"

I tried finding kinetic and gravitational energy, but then I realized I don't have any mass to use it with (1/2)*m*v^2 and m*g*h

How can I solve this problem without guessing and checking?

Use the appropriate kinematics formula for uniform acceleration.

The second post in the link below should help.
https://www.physicsforums.com/showthread.php?t=110015

[Hint: you'll have to solve for t once you have the right formula.]

 

[Paincake]

I've seen online a kinematic equation for final velocity, which is

Vf² = Vi² + 2(a)(d),

but my teacher has not introduced it. Where does this equation come from?

EDIT:
Nevermind, I see it now.

I solved for t using vf = vi + at --> (vf-vi)/a = t
And I know distance = v*t and average velocity = (vi+vf)/2

So it was a matter of simplifying d = (vi+vf)/2 * (vf-vi)/a to d = (vf^2+vi^2)/2a and solving for vf.

Thanks for the help.

 

[collinsmark]

I've seen online a kinematic equation for final velocity, which is

Vf² = Vi² + 2(a)(d),

but my teacher has not introduced it. Where does this equation come from?

That's not the equation you want to use for this problem. Try to find a formula that has initial velocity (not final velocity), distance, acceleration, and time.

But I'll tell you where that equation comes from anyway, if you're curious. It comes from combining two of the other kinematics equations for uniform acceleration.

Start with

x = x_0 +v_0 t + \frac{1}{2}at^2

Now modify some variables, using d = x - x₀. And instead of calling the initial velocty "v₀", let's call it "v_i" instead. So now we have,

d = v_i t + \frac{1}{2}at^2

Don't forget about that equation, we'll come back to it in a second. But first let's look at a different kinematics equation for uniform acceleration:

v_f = v_i + at

Rearranging that equation, we have

t = \frac{v_f - v_i}{a}

Now let's substitute that into the modified first equation above.

d = v_i \frac{v_f - v_i}{a} + \frac{1}{2} a \left( \frac{v_f - v_i}{a} \right)^2

Expanding a little gives us

d = \frac{v_i v_f - v_i^2}{a} +\frac{1}{2}a \left( \frac{v_f^2 - 2v_f v_i + v_i^2}{a^2} \right)

= \frac{v_i v_f - v_i^2}{a} + \frac{v_f^2 - 2v_f v_i + v_i^2}{2a}

Multiplying both sides of the equation by 2a gives,

2ad = 2v_i v_f - 2v_i^2 + v_f^2 - 2 v_i v_f + v_i^2.

And simplifying the right side of the equation produces

2ad = v_f^2 - v_i^2.

Adding v_i² to both sides gives

v_i^2 + 2ad = v_f^2.