Say you have a missile that is launched from the ground at $t = 0$, with an initial velocity of $80.1 ms^{-1}$ at an angle of $57.3$ degrees. At some time it is displaced $200m$ vertically and has a velocity of $50ms^{-1}$ at an angle of $30$ degrees.

Why can I not find the correct time at which the missile reaches $200m$ vertical displacement using $v = u + at$? I attempted to fill it in as: $$t = \frac{50sin30 - 80.1sin57.3}{-9.8}$$

However this got me $4.32s$ when the correct answer is $9.42$, using $s = ut + \frac{1}{2}at^2$. Why does my way not work?

  • $\begingroup$ Are you sure that 9.42 is the correct answer? The quadratic has two roots. One corresponds to an altitude of 200m while going up, the other while coming back down ($v_y<0$). $\endgroup$ – amd May 8 '16 at 16:39
  • $\begingroup$ Yeah its from an A-Level past paper mark scheme. $\endgroup$ – Tiernan Watson May 8 '16 at 16:44
  • $\begingroup$ Never mind. My mistake. "when at 200m for the second time". So to use my equation, you would have to make the velocity negative. But you are not given its velocity when coming back down, so probably better to use the quadratic to be safe. $\endgroup$ – Tiernan Watson May 8 '16 at 16:46
  • $\begingroup$ That makes sense now. You don‘t have to use the quadratic, though. The velocity should have an angle of $-30$ degrees when it passes 200m the second time. The $x$-component hasn’t changed, and the $y$-component will have the same magnitude as on the way up by symmetry (and conservation of energy). $\endgroup$ – amd May 8 '16 at 16:47

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