Q2.5 Steep and shallow trajectories (S)

A ball is launched at a speed of 10 ms-1. It lands 5 m away. Find the two possible values of the angle, θ, which the initial trajectory makes with the horizontal. [Take g=10 ms-2.]

Sketch (on the same figure) the trajectories associated with these two launch angles, and explain which (if either) takes the longer time to complete.

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Hint

To pick up the two solutions you’ll need to remember that
sinφsin(180-φ)
 

Solution

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Solution

The time of flight follows from
\[ 0=y-y_0= v_{y0} t + \frac{1}{2} (-g) t^2 \hspace*{0.5cm} \mbox{implying} \hspace*{0.5cm} t_f= \frac{2v_{y0}}{g} \]
The range is then
\[ R=x-x_0= v_{x0} t_f = \frac{2v_{x0}v_{y0}}{g}=\frac{v_0^2\sin(2\theta)}{g} \]
Thus
\[ \sin(2\theta) = \frac{Rg}{v_0^2} = \frac{5 \times 10}{100} = \frac{1}{2} \]
This equation has one obvious solution, namely 2θ=30. But there is another one: since sinφsin(180-φ), the choice 2θ=180-30=150 is also a solution. We conclude that the solutions for θ are
θ=15,75
The time of flight reflects the y component of the initial velocity. It is longer for the trajectory with the bigger launch angle.
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