A particle is projected from O with speed 8 ms $^{-1}$ at an angle $\theta$ above the horizontal and passes through the point $A(8;\ -1.81)$ , distances in metres. The two possible values of $\theta$ are $45^{\circ}$ and $32.2^{\circ}$ . Which value gives the minimum time taken to reach A, and why?

32.2^{\circ}

, because

8\cos\theta

is larger, so the time

t=\sec\theta

to cover the 8 m is smaller.

45^{\circ}

, because

45^{\circ}

always gives the fastest projectile flight.

45^{\circ}

, because the vertical component

8\sin\theta

is larger, so the particle falls to the level of A sooner.

DBoth give the same time, because both reach the same point A from the same point O.

Explanation

The horizontal motion is at the constant speed

8\cos\theta

, and A is 8 m horizontally from O, so the time to reach A is

t=\dfrac{8}{8\cos\theta}=\sec\theta

. That increases with

\theta

, so the flatter path is the quicker one:

\sec32.2^{\circ}=1.18

s against

\sec45^{\circ}=1.41

s. The angle

45^{\circ}

maximises range on level ground, which is a different question, and the two paths reach A at different times.

Derived from ZIMSEC Mathematics 9164/4 Paper 4, November 2004, Q14

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