Solve Speed Problem: Plate Reaches 40m Height & 70m Length

[Kahsi]

Hi.

A man is throwing a plate in the air. The plate reaches the height of 40m and the length of 70m. What was the speed of the plate?

This is what I've done

v_y:

mgh = \frac{mv^2}{2} => v = \sqrt{2gh} = \sqrt{2*9,82*40} = 28m/s


v_x:

v = v_0 + at => t = 28/9,82 = 2,85s

s = vt => v = s/t = 70/2,85 = 24,56m/s

                       /|
                     /  |
                   /    | v_y = 28m/s
           v_z   /      |
               /        |
             /          |
           /            |
         /______________|
         v_x = 24,6m/s

v_z = \sqrt{v_y^2+v_x^2} = \sqrt{24,6^2 + 28^2} = 37m/s

But the answer should be 31m/s. What am I doing wrong?

Thank you.

 

[krab]

Initial v_y is 28 m/s, but final is -28m/s, so total change is 56m/s and you've calculated t incorrectly; it is actually twice the value you gave.

 

[Kahsi]

Thank you for the reply krab.

I have now another question.

We know that
mgh = \frac{mv^2}{2} => h = \frac{v^2}{2g}
And we also know that s = v_0t + \frac{at^2}{2}
When it's a free fall v_0 = 0. That gives us
s = -\frac{gt^2}{2}

So what's the difference between s and h?

Thank you.

 

[James R]

Your expression for h is in terms of the speed v when the object hits the ground.

The object's speed at time t is

v = gt

so t=v/g

Putting this into your expression for s, we get:

s = -\frac{gt^2}{2} = -\frac{g}{2}\frac{v^2}{g^2} = -\frac{v^2}{2g} = -h

So, as you can see, h and s are the same, apart from the negative sign. The negative sign comes from the fact that in one case you've measured positive distances downwards, and in the other positive distances are upwards.