enter image description here

Hi everyone,

How do I approach questions like part (ii)?

From what I understand now, I believe the shortest method is that as long as the ANGLE of the relative velocity of B with respect to velocity of A is = Angle of thetha which in this case is arctan(50/150)?

Is this correct?

Another method I tried was to find the time taken for projectile and intercepter to reach the same x coordinate and use this to obtain the time in terms of thetha. Then I equated the y coordinate using this time in terms of thetha but I ended up being unable to solve them with 10sin - 10/3cos sin + cos = sqrt(3) + 3

What other ways are possible as well?

Thank you

  • $\begingroup$ The motion of such projectiles is never linear, but parabolic $\endgroup$
    – Fred
    Commented Apr 27, 2019 at 9:35

2 Answers 2


First, use consistent angles. Define $\theta^*$ to be the supplementary angle to the one illustrated.

For an intercept to occur, $x_A(t) = x_B(\theta^*,t)$ and $y_A(t) = y_B(\theta^*,t)$ for some $\theta^*$ and time $t$.

Taking $A$ as the origin -

$$x_A = 30 cos(30) t \qquad\qquad\ y_A=-1/2gt^2 +30sin(30)t\qquad$$ $$x_B = 50 cos(\theta^*) t + 150 \qquad y_B = -1/2gt^2 + 50sin(\theta^*)t - 50$$

Solve for the angle and time, and don't forget to change the angle back to it's supplement.

  • $\begingroup$ isn't xB=150 - 50cos(θ∗)t ? and i think this is my 2nd method which resulted in 10sin - 10/3cos sin + cos = sqrt(3) + 3 $\endgroup$ Commented Apr 29, 2019 at 4:40
  • $\begingroup$ @OngCheeWei Thetastar is the supplementary angle to theta, so the cosine's sign changes. Yours is correct for the angle shown in the illustration. But often you have a lot of angles, and it is good to get into the habit of running all of them ccw from the x axis in the x - y plane. $\endgroup$
    – Phil Sweet
    Commented Apr 29, 2019 at 10:14

You can write the parametric equation of trajectories:

$$y= -1/2gt^2+v_{y_{0}}t+y_0 \quad and\quad x= v{x_{0}}t+x_{0}$$

for the A it is,

$$y_A=-1/2gt^2+1/2*30*t+50 $$ And for B,we just use sin(theta)

$$ y_B=-1/2gt^2+50*sin(\theta)t+0 $$

We eliminate t and find the tetha.

Then we plug in the below equations to find t and x.

$$ x_B=150- 50 cos(\theta)t$$ And $$ x_A = 30 cos(30)t+0$$ $$150-20cos(\theta)t =0 \quad cos(\theta)t=7.5$$

  • $\begingroup$ Tetha aka theta? $\endgroup$
    – Solar Mike
    Commented Apr 27, 2019 at 7:14
  • $\begingroup$ Thanks Solar Mike. I corrected the spelling. $\endgroup$
    – kamran
    Commented Apr 27, 2019 at 7:20
  • $\begingroup$ no, kamran, this isn't sufficient. To effect an intercept, both the x and y components have to match at the same time. The variables available are theta and time. You can't eliminate time, you need it. The yb formula is missing a 2. Lastly, you need to make the trig consistent with the angles as drawn or redefine the angles to be consistent. $\endgroup$
    – Phil Sweet
    Commented Apr 27, 2019 at 16:50
  • $\begingroup$ Phil Sweet, it is obvious that x, y, t should be the same. We have 3 unknowns and 3 equations. I will finish it if OP asked. $\endgroup$
    – kamran
    Commented Apr 27, 2019 at 16:57
  • $\begingroup$ Three edits and it's still not right. You can't eliminate t and find theta. You have only one formula that looks like 50sin(theta)t = 15t+50. $\endgroup$
    – Phil Sweet
    Commented Apr 27, 2019 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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