# Thread: determine force and height.

1. ## determine force and height.

A projectile of mass 1.050 kg is shot straight up with an initial speed of 24.0 m/s.

(a) How high would it go if there were no air friction?
in meters

(b) If the projectile rises to a maximum height of only 23.2 m, determine the magnitude of the average force due to air resistance.
in Newtons

2. Originally Posted by rcmango
A projectile of mass 1.050 kg is shot straight up with an initial speed of 24.0 m/s.

(a) How high would it go if there were no air friction?
in meters

(b) If the projectile rises to a maximum height of only 23.2 m, determine the magnitude of the average force due to air resistance.
in Newtons
(a)

$\displaystyle (v_{f})^2=(v_{i})^2+2ad$

$\displaystyle 0=24.0^2+2(-9.81)d$

$\displaystyle 0=576+(-19.62)d$

$\displaystyle -576=-19.62d$

$\displaystyle \frac{-576}{-19.62}=d$

$\displaystyle 29.36=d$

not so sure about b but i think you could factor the air resistance into acceleration because its average force so ill try it that way

$\displaystyle (v_{f})^2=(v_{i})^2+2ad$

$\displaystyle 0=576+2a(23.2)$

$\displaystyle -576=46.4a$

$\displaystyle -12.41=a$

$\displaystyle -12.41-(-9.81)=-2.6$

$\displaystyle f=ma$

$\displaystyle f=(1.050)(-2.6)=-2.73n$

3. Originally Posted by rcmango
A projectile of mass 1.050 kg is shot straight up with an initial speed of 24.0 m/s.

(b) If the projectile rises to a maximum height of only 23.2 m, determine the magnitude of the average force due to air resistance.
in Newtons
Here's a more systematic approach to b) since you are doing energy problems.

We know there is air resistance, so we know that there are non-conservative forces lurking about. Thus
$\displaystyle W_{nc} = \Delta E$

The force due to air resistance (F) is always in the opposite direction of the displacement, so
$\displaystyle W_{nc} = \vec{F} \cdot \vec{s} = Fs \cdot cos(180^o) = -Fs$

So
$\displaystyle -Fs = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 + mgh - mgh_0$

Call upward positive and set the zero point for the gravitational potential energy at the point where the projectile is shot upward. (Typically I choose the 0 point for GPE at the lowest point in the diagram. That way I don't have to mess around with negative h values.) So $\displaystyle h_0 = 0~m$, and since the projectile is going to the max height, v = 0 m/s. Also note that $\displaystyle s = h$ in this coordinate system. Thus
$\displaystyle -Fh = - \frac{1}{2}mv_0^2 + mgh$

$\displaystyle F = \frac{mv_0^2}{2h} - mg$

You can plug the numbers in.

-Dan

4. Thankyou for the work here, I didn't realise that i double posted this problem.

I appreciate the solutions.