Kinematics - When projectile lands at an angle

5,459

As always, a diagram clears things up quickly:

enter image description here

We are going to assume that there is no lift on the arrow (which is wrong - arrows do not fly like regular projectiles but that is not the point of this question).

From thie diagram we can see the approach we need to take (I am not going to show the details of the steps, just give you some direction).

  • the time $t$ taken to fly 60 m is calculated from the distance $d$ (60 m) and the horizontal velocity $v_h$ (unknown): $t = \frac{d}{v_h}$
  • During that time, the arrow picks up vertical velocity - $v_v = g t$
  • the ratio of vertical and horizontal velocities is $\frac{v_v}{v_h}=\tan \theta$ - this allows us to write $v_v$ as a function of $v_h$

Combining the above, you can eliminate $v_v$ from your equations and you will be left with an expression for $v_h$ in terms of $g$, $d$ and $\theta$.

Share:
5,459

Related videos on Youtube

Donald Dang
Author by

Donald Dang

Updated on August 01, 2022

Comments

  • Donald Dang
    Donald Dang over 1 year

    Question:

    You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground $60 m$ away, making an angle of 3 degrees with the ground. How fast was the arrow shot?

    Attempt at solution:

    I split the projectile into $x$ and $y$ components, where $dx = v_1t$ and $v_1 = v_0 +at$. Using the $dx$ equation, I isolated for $t$ and got $\dfrac{60}{v_1}$. I took the vertical and horizontal components for the angle the arrow is landing at which is $-\tan 3°$. Since I assumed $v_0 = v_1$, I get $t= \dfrac{60}{-\tan 3°}$

    From here, I don't know where to go. In the solutions it says that $-g/t/(60/t)= -\tan3°$, but I don't know how and where they got it from. Can anyone help me?

    • Abhijeet
      Abhijeet over 8 years
      Tilt the whole system 3 degrees clockwise, now the arrow is shot at 3 degrees from the horizontal and lands flat at ground. Now using Range, R = (u^2)sin2theta/g, solve.