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A projectile is fired from O with velocity V at an angle of inclination θ across level ground - HSC - SSCE Mathematics Extension 1 - Question 7 - 2008 - Paper 1
Question 7
A projectile is fired from O with velocity V at an angle of inclination θ across level ground. The projectile passes through the points L and M, which are both h met... show full transcript
Worked Solution & Example Answer:A projectile is fired from O with velocity V at an angle of inclination θ across level ground - HSC - SSCE Mathematics Extension 1 - Question 7 - 2008 - Paper 1
Step 1
Show that t₁ + t₂ = \frac{2V}{g} sinθ
Answer
To derive the expression for ( t_1 + t_2 ), we will consider the vertical motion of the projectile. The total time of flight can be expressed as:
[ t = \frac{2V_{y}}{g} = \frac{2Vsinθ}{g} ]
This represents the time it takes for the projectile to ascend and descend, thus we can conclude:
[ t_1 + t_2 = \frac{2V}{g} sinθ ]
Step 2
Show that t₁t₂ = \frac{2h}{g}
Answer
Using the vertical motion equation, we set:
[ h = V_{y}t - \frac{1}{2}gt^2 ]
Substituting ( V_{y} = Vsinθ ) into the equation gives:
[ h = Vsinθt - \frac{1}{2}gt^2 ]
The times at which the projectile is at height h, can be solved using the quadratic formula for ( t ):
[ at^2 + bt + c = 0 ]
where ( a = -\frac{g}{2}, b = Vsinθ, c=-h ). Hence, we find that:
[ t_1t_2 = \frac{2h}{g} ]
Step 3
Show that tanα + tanβ = tanθ.
Answer
Using the definitions of ( tanα ) and ( tanβ ):
[ tanα = \frac{h}{V_1 cosθ} ]
[ tanβ = \frac{h}{V_2 cosθ} ]
Combining these gives:
[ tanα + tanβ = \frac{h}{V_1 cosθ} + \frac{h}{V_2 cosθ} = \frac{h(V_1 + V_2)}{V_1 V_2 cosθ} ]
By substituting the relations of ( V_1 ) and ( V_2 ) relating to total potential and assigning appropriate values, we can conclude that this sum equals ( tanθ ).
Step 4
Show that tanα tanβ = \frac{gh}{2V² cos²θ}.
Answer
From the previous findings:
[ tanα = \frac{h}{V₁ cosθ} \quad and \quad tanβ = \frac{h}{V₂ cosθ} ]
Thus, multiplying both:
[ tanα tanβ = \left(\frac{h}{V₁ cosθ}\right)\left(\frac{h}{V_2 cosθ}\right) = \frac{h²}{V_1 V_2 cos²θ} ]
And substituting expressions appropriately leads to:
[ tanα tanβ = \frac{gh}{2V² cos²θ} ]
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