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

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

 

[asteriatic]

I would approach this problem by using the equations of motion for a body in free fall. The initial velocity of 20 m/s and the displacement of 300 m can be used to calculate the time required to fall using the equation: s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s^2 on Earth), and t is the time.

By substituting the given values into the equation, we get: 300 m = (20 m/s)t + (1/2)(9.8 m/s^2)t^2. This is a quadratic equation that can be solved using the quadratic formula. The resulting positive root will give us the time required to fall 300 m.

Alternatively, we can also use the equation: v = u + at, where v is the final velocity (which will be 0 when the body hits the ground) and all other variables are the same as before. By substituting the given values, we get: 0 = 20 m/s + (9.8 m/s^2)t. Solving for t, we get the same result as before.

Therefore, by using the equations of motion, we can accurately calculate the time required for the body to fall 300 m without having to guess or check.