Q1.8 Asking yourself if it makes sense (K)
A projectile is launched with speed v0 at angle θ above the ground. If the effects of air-resistance are ignored its range (the horizontal distance to the point at which it returns to the ground) is given by![\[ R= \frac{2 v_0^2 \sin \theta \cos \theta}{g} \]](mastermathpng-0.png)
- Check the consistency of the units.
- Check that the equation is consistent with the special cases θ=0, and θ=90∘.
- Think about the qualitative dependence of the result upon g.
- Think about why the result grows with the square of the launch speed.
Hint
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Hint
In the last bit you need to recognise that the launch speed affects the range in two different ways: through the time the projectile is in the air; and through the horizontal speed at which it travels during this time.Solution
Reveal
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Solution
- The units of the left hand side are R→m. The units of
the rhs are
The two sides match up; so the equation passes this test (Key Point 0.2, Guideline 0.9).
- If the projectile is launched horizontally (θ=0) we expect it will hit the ground immediately, giving zero range. Since sin0=0, the range formula is consistent with this special case (Guideline [GL-specialcase in Node probsolving]). If the projectile is launched vertically (θ=π/2) it will return to the same point on the ground; again giving zero range. Since cosπ/2=0, the range formula is consistent with this special case too.
Qualitatively we’d expect the range to be bigger the smaller is g: after all in the absence of any gravitational acceleration there would be nothing to bring the projectile back to the horizontal at all!
Aside: You may be bothered by being asked to think of g as a variable. Isn’t g=9.8ms-2? Indeed, if we are dealing with a projectile at the earth’s surface we would assign this value to g...at the end of the calculation. But the argument and the result should hold for general g. So we are quite at liberty to ask that it hold for any g we like. This is one of the reasons for Guideline 0.6.
- This bit is a little harder, and (if you have not had much experience of kinematics) may not be clear until we have covered the material of S1. The point is that an increase in v0 increases the range through two distinct mechanisms. First, it increases the horizontal component of the launch velocity; and second it increases the vertical component of the launch velocity and thence the time the projectile gets to travel horizontally before it is stopped by its reencounter with the ground.