Question

At time t = 0 a small ball is projected from point A with a velocity of 200 ft/sec at the 6060^{\circ} angle. Neglect atmospheric resistance and determine the two times t1t_1 and t2t_2 when the velocity of the ball makes an angle of 4545^{\circ} with the horizontal x-axis.

Solution

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

v=(v0cosθ, v0sinθgt)\vec v=(v_0\cos\theta,\ v_0\sin\theta-gt)

We are looking for a solution of the following equation.

tan(±45)=vyvxvx=±vy\tan (\pm45) = \frac{v_y}{v_x}\rightarrow v_x=\pm v_y

±v0cosθ=v0sinθgt\pm v_0\cos\theta = v_0\sin\theta - gt

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

t=v0g(sin60±cos60)=200232.174(3±1)t=\frac{v_0}{g}(\sin 60 \pm\cos 60)=\frac{200}{2\cdot 32.174}(\sqrt 3\pm 1)

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

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