Anonymous
Anonymous asked in Science & MathematicsPhysics · 6 months ago

# When an object is thrown in the air and falls back down we use this quadratic equation:--- >>>> (see more below)--- >>>?

------ >>>>> (h = -4.9t^2 + v0t + h0) to calculate speed, time, height, distance etc. What is the proof for using this formula(h = -4.9t^2 + v0t + h0)? I admit I know very little about physics and I would need a very long explanation.

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• No proof is needed.

The evidence is that the acceleration of gravity is a constant value.

This value is approximately 9.8 m/s^2.

This has been found over and over again by experiment but it is only a close approximation.

It is almost correct for large dense objects but it is completely wrong for things like a balloon.

Having established that the rest is an automatic consequence.

The displacement is the integral of the velocity.

The velocity is the integral of the acceleration.

ie a = g = - 9.8

v = vo + gt = vo - 9.8 t

s = integral (v) = so + vo t + 1/2 g t^2

= so + vot - 4.9 t^2 ( approximately.... remember that the exact value of the acceleration of gravity is not PRECISELY 9.8 )

• A derivation which you can undersand depends on your starting point - what background (in maths and physics) you already have.

For example do you know how to use acceleration and time to find a change in velocity? Do you know about sign conventions for acceleration and velocities? Do you understand what 'g' (acceleration due to gravity) is?

Without knowing your starting point, it is impossible to give a good explanation here.

You could try the 3 videos in the link - but note the symbols used are different to the ones you are using (e.g. initial velocity is 'u').

• Let's say I toss a ball up and it leaves my hand with speed +v0 where the "+" indicates it is moving upward. I know there is a net force acting on the ball namely

F = -m*g where m = mass of ball and g = 9.8 m/s^2 and the "-" means it is acting downward

By Newton's laws of motion I can write F = -m*a = -m*g or a = g

Now a = rate of change of speed or velocity over time = dv/dt so

g = dv/dt --> and v = change in position over change in time = dh/dt so

g = dv/dt = d(dh/dt)/dt = d^2h/dt^2

Integrate once gt = dh/dt + constant. we know that at t = 0 dh/dt = +v0 so

gt = dh/dt + v0 --> integrate again

(1/2) gt^2 = h +v0*t + constant --> set constant = h0 = initial position at t = 0

(1/2) gt^2 = h +v0*t + h0 --> -h = -1/2 gt^2 +v0*t + h0

• I don't have a proof, but I do have a derivation based on the physics.

Let's assume a rock is tossed with initial speed U at angle theta above the ground and launched from height h above the ground. We want the vertical height y(t) above the ground at any time t since being tossed.

So first of all it's at y(t = 0) = h before we toss it. And that's our first term in the equation. Then we fling it with Ux = U cos(theta) X speed and Uy = U sin(theta) Y speed.

And that means we have y(t) = h + Uy t as the height after t time of vertical flight at speed Uy. But, a big BUT, there is the gravity of the situation. So that rock does not continue unabated at Uy, it starts to slow down due to gravity.

And that's our next term, the gravity term 1/2 gt^2 which if you will recall is the distance traveled while in a constant acceleration over time t. But this gravity distance is actually a loss of distance since the rock is decelerating at g the gravity rate.

And that means we need to subtract 1/2 gt^2 from y(t) = h + Uy t to get y(t) = h + Uy t - 1/2 g t^2, which is your h = -4.9t^2 + v0t + h0, note that 1/2 g = 4.9. QED.