Question

derive expression for time of flight

Answer 1

Answer:

The time of flight of an object, given the initial launch angle and initial velocity is found with: T=2visinθg T = 2 v i sin ⁡ . The angle of reach is the angle the object must be launched at in order to achieve a specific distance: θ=12sin−1(gdv2) θ = 1 2 sin − 1 ⁡ ( gd v 2 ) .

Answer 2

$\underline{\bigstar{\sf\ Answer:}}$

Consider the above diagram.

Motion considered in y - direction.

O -------> Q

uy = usin θ

ay = - g

vy = 0

By 1st equation of motion

v = u + at

0 = usin θ + ( -g × t)

- usin θ = -gt

$t = \frac{u \sinθ }{g}$

Total time taken = 2t

➡️ $T\: = \frac{2u \sinθ}{g}$

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