Question

A projectile is launched vertically with a velocity of 30 mxs . How long will it take to return to the original launch position?

Answer 1

Answer: If the projectile is launched vertically, then you only aply velocity on the y axis.

The velocity 30 m/s is the initial velocity and if it is fired on the ground, then the initial position is 0m (this doesn't really matter), and now, let's analyze the forces on the projectile.

Once it is fired, the only force acting on the projectile is the force of gravity, and we know that the gravity acceleration is $-g = -9.8\frac{m}{s^{2} }$, where the negative sign is there because this acceleration points downward.

so A(t) =  $-9.8\frac{m}{s^{2} }$

For the velocity, we need to integrate the acceleration in the time, this is:

$v(t) = -9.8\frac{m}{s^{2} }*t + v0$

where v0 is the initial velocity, in this case 30m/s

And for the position, we need to integrate again, so:

$r(t) = -4.9\frac{m}{s^{2} }*t^{2} + 30\frac{m}{s} *t + r0$

where r0 is the initial position, in this case 0m.

now the question is "How long will it take to return to the original launch position?"

So now we need to find the time in which r(t) is zero again (so the projectile is in the ground again

so r(t) = 0 = $r(t) = -4.9\frac{m}{s^{2} }*t^{2} + 30\frac{m}{s} *t + r0$

this is: r(t) =  t*(-4.9*t + 30) = 0

so is easy to see that t = 0 (because it is fired in the ground) is a solution, but is not the one that we are looking for,

so we only look inside the parentheses:

-4.9*t + 30 = 0

t = 30/4.9 = 6.12 s

So 6.12 seconds after the projectile is fired, it returns to the ground, or the original launch position.