Question

# At time t = 0 a small ball is projected from point A with a velocity of 200 ft/sec at the $60^{\circ}$ angle. Neglect atmospheric resistance and determine the two times $t_1$ and $t_2$ when the velocity of the ball makes an angle of $45^{\circ}$ with the horizontal x-axis.

Solution

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The velocity vector of the ball is $\vec v$. The tangent to the trajectory looks in the same direction.

$\vec v=(v_0\cos\theta,\ v_0\sin\theta-gt)$

We are looking for a solution of the following equation.

$\tan (\pm45) = \frac{v_y}{v_x}\rightarrow v_x=\pm v_y$

$\pm v_0\cos\theta = v_0\sin\theta - gt$

Solving this equation for $t$ and using the given values leads us to the final answer.

$t=\frac{v_0}{g}(\sin 60 \pm\cos 60)=\frac{200}{2\cdot 32.174}(\sqrt 3\pm 1)$

$\boxed{t=8.491,\ 2.275\ \mathrm{s}}$

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