0
$\begingroup$

I have a physics problem related to projectile motion, I am having difficulties figuring out how to find the initial speed in the $x$ and $y$ directions, as well of the height of the drop. I've tried reading the text book several times but I can't figure out of to do it. We're using the University Physics with Modern Physics 14th Ed by Young and Freedman, if that matters.

The picture below is a paint model of known variables

I know that

  • $x_0 , y_0$ = initial position x and y direction
  • $v_{0_x} , v_{0_y}$ = initial speed x and y direction
  • $t$ time
  • $a_x = 0$
  • $a_y = -g$ $$x = x_0 + v_{0_x}\cdot t, v_x = v_{0_x}$$ $$y = y_0 + v_{0_y}\cdot t - \frac12g\cdot t^2, v_y = v_{0_y} - g\cdot t$$

also that $y = 6.74 cm$ when $v_y = 0$

Can anyone point me in the right direction? No need to account for wind resistance.

Known variables

$\endgroup$
0
$\begingroup$

As I mentioned in a comment, set $v_{0_x}=v_0\cos\theta$ and $v_{0_y}=v_0\sin\theta$.

To guide you: set $t_m$ the time at which the projectile is at the top of the parabola with $v_y(t_m)=0$ you find an expression for $t_m$ then inject it in $y(t_m)=h$ with $h=6.74$ cm. Solve the equation for $v_0$

This gives you an expression for $v_0$, and I would strongly suggest that you set the origin to $(x(0),y(0))$ so that you don't have to handle your useless variables $x_0$ and $y_0$.

At this point you have $v_0$, find the time $t_f$ at which the position in x of the projectile is $L$ with $L=1.06$ m. Then calculate $y(t_f)$ and you should find something $<0$, which corresponds to the position where the projectile hit the ground with respect to its starting position.

$\endgroup$
  • $\begingroup$ thanks! once I figured out how to get to $v_0$ it was alot easier to figure out the rest $\endgroup$ – Iltharion Mar 1 '16 at 11:51
  • $\begingroup$ how would I og about if I needed to find $v_o$, if I don't know the $v(t) = 0$ but I know how far it flies in the x-direction and and how far up on a wall it lands? say it hits a wall $x = 18m$ away from $x_0$ and $y = 8m$ above $y_0$ $\endgroup$ – Iltharion Mar 1 '16 at 13:26
0
$\begingroup$

Your x speed is constant because there is no force in this direction. With the formula of position according to initial speed and initial position, you'll find out. Tip: dont forget to use the correct sign

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.