Question

player throws a ball with velocity of 49m/s at angle of 45˚ with the horizontal. How long later,

the ball would hit the ground
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Answer 1

Answer:

initial velocity (u) =49 m/s

angle of projection (x) =π/4

$vertical \: motion \\ y - component \: of \: u = u \sin(x) = \frac{49}{ \sqrt{2} } m {s}^{ - 2} \\ displacement(s) = 0 \\ time \: of \: flight \: = \frac{2u \sin(x) }{g} \\ t = \frac{2( \frac{49}{ \sqrt{2} } ) }{9.8} \: \\ t = \frac{49 \times \sqrt{2} \times10}{98} \\ t = 5 \sqrt{2} s$