I have the question "An arrow is fired from a height h meters at a speed of 55 m/s and at an angle of 5 degrees above the horizontal. The arrow hits a target at a horizontal distance of 60 m from where it was fired and at a vertical height of 1.2 m.

a) show that the time for which the arrow is in the air before it hits the target is 1.10 seconds (to 3.s.f).

b) Find the value of h. "

Here is my attempt for part a) :

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However I do not get the answer of 1.10 seconds what have I done wrong here ?

Here is my attempt for part b) is this correct ?

enter image description here


Try drawing a diagram and indicate how the velocity is split into two components.

Part a)
If you fire the target at an angle and it stops 60 meters away then you need only find the horizontal component of velocity and divide it into the horizontal distance away from the starting point (as there is no acceleration in x direction.
$s=v cos(\theta) t $
Now try use this for your values as per question.

Part b)
If $d$ is the finishing height and $h$ is the starting height.
Then the formula becomes:
$d=v sin(\theta) t -\frac{1}{2}gt^2 + h$
Where we know the total time from part a. So we will have only one unknown which is h.
Use your values in this equation to solve and I also recommend plotting the graph using graphing software to visualise the trajectory .

Remember d is the final VERTICAL HEIGHT. You're saying the final veryical height is 60 when in fact that's the distance along the X axis.

  • $\begingroup$ I got 0.107 metres for part b) is this correct ? :) $\endgroup$ – Dan Dec 5 '17 at 0:01
  • 1
    $\begingroup$ @Dan my answer was 1.92 . 1.2 = 55sin(5)(1.1) -1/2 (9.8)(1.1)^2 + h. Then solve for h. $\endgroup$ – Matthew Dec 5 '17 at 20:21

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