Optimal angle for maximum displacement

Optimal angle for maximum displacement

First look at the total horizontal displacement (from initial to final position)
***not only half the distance***

Since the horizontal velocity is constant you can write the total horizontal displacement as follows:

$\varDelta r_x = v_xt$

We now want to modify this equation so that we have the total displacement at the final time. In order to do this, we take advantage of the fact that the time from the initial position to the highest point is equal to the time from the highest point to the final position:

$t_{total} = 2t_{up}$

Also, since we know that at the highest point the vertical velocity is equal to zero, we can isolate time from the following equation:

$v_m = v_{yi} + at_{up}$

$t_{up} = \frac{v_m - v_{yi}}{a}$

Since vertical velocity at maximum height is equal to zero we can write

$t_{up} = \frac{-v_{yi}}{a}$

But this is only half the time! Therefore, when we substitute this equation into the equation for horizontal displacement, we have to multiply it by 2.

therefore we end up with

$\varDelta r_x = \frac{v_{xi} (-2v_{yi})}{a}$

In order to continue with this problem, we need to look at $v_{xi}$ and $v_{yi}$ in terms of $v_i$

$v_x = v_i cos\Theta$
$v_y = v_i sin\Theta$

We now have

$\varDelta r_x = \frac{v_i cos\Theta (-2v_i sin\Theta)}{a}$

$\varDelta r_x = \frac{-v_i^2 2cos\Theta sin\Theta}{a}$

This equation is also given in your text book.

But where does that leave us?

If you remember your identities (or have looked them up)

$2cos\Theta sin\Theta = sin2\Theta$

We can therefore write

$\varDelta r_x = \frac{-v_i^2 sin2\Theta}{a}$

Horizontal displacement is therefore a function of gravity (which we cannot change) initial velocity (we want the optimal angle for all magnitudes of initial velocity) and $sin2\Theta$ (we need to find a theta which will maximize horizontal displacement!).

Now, you need to remember that $sin()$ can vary between values of -1 to values of 1.

To answer question 4
think of what the value of $sin2\Theta$ needs to be in order to maximize horizontal displacement and solve for $\Theta$

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Tags: MVS330