At time t=0, a small ball is projected from point A with a velocity of 60 m/s at $60 °$ angle with horizontal. Neglect atmospheric resistance and determine the two times $t_{1}$ and $t_{2}$ when the velocity of the ball makes an angle of $45 °$ with the horizontal x-axis.

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$t_{1} = 2.2 s and t_{2} = 8.2 s$

$t_{1} = 1.2 s and t_{2} = 8.2 s$

$t_{1} = 2.2 s and t_{2} = 4.2 s$

$t_{1} = 1.2 s and t_{2} = 4.2 s$

answer is A.

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Detailed Solution

$When velocity vector makes an angle 45 ° with horizontal, v_{y} = \pm v_{x} = \pm u_{x} = \pm 60 \cos 60 ° = \pm 30 m/s$

Now, $v_{y} = u_{y} + a_{y} t$

$or, t = \frac{v_{y} - u_{y}}{a_{y}} = \frac{(\pm 30) - 60 \sin 60 °}{- 10} = 3 (\sqrt{3} \pm 1) s ∴ t_{1} = 2.2 s and t_{2} = 8.2 s$

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