Question

a projectile is thrown in the upward direction making an angle of 60 with horizontal direction with a velocity of 140 m/s . then the time after which its inclination with the horizontal is 45 degree, is?​

Answer 1

Answer:

here the anser of your question

Answer 2

Answe

$velocity \: of \: projectile \: = 147 \frac{m}{s } \\ \\ angle \: of \: projection \: = 60 \: degree \: \\ \\ \\ let \: the \: time \: which \: is \: taken \: by \: projetive \: from \: 0 \: to \: a \: be t \: and \: its \: velocity \: is \: v \: at \: point \: a \: where \: the \: drection \: is \: b \: = 45 \\ \\ horizontal \: component \: of \: velocity \: remains \: constant \: during \: the \: projectile \: motion. \\ \\ v \: cos \: 45 \: degree \: = v \: cos \: 60 \: degree \: \\ \\ v \times \frac{1}{ \sqrt{2} } = 147 \times \frac{1}{2} \\ \\ \frac{147}{ \sqrt{2} } {ms}^{ - 1} \\ \\ for \: vertical \: motion \: \\ \\ v y = uy - gt \\ \\ vs \: in \: 45 \: degree \: = 45 \: in \: 60 \: degree \: - 9.8t \: \\ \\ \frac{147}{ \sqrt{2} } \times \frac{1}{ \sqrt{2} } = 147 \times \frac{ \sqrt{3} }{2} - 9.8t \\ \\ 9.8t \: = \frac{147}{2} ( \sqrt{3 \: - 1) } \\ \\ t = 5.49s$

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