A projectile is released with a speed u at an angle \(\theta\) to the horizontal. With the aid of a diagram, show that the time of flight is equal to \(\frac{2uSin\theta}{g}\), where g is the acceleration of free fall.
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Correct Answer: Option n
Explanation:
Vertical component velocity Vy = Usin\(\theta\)
Time to reach maximium height = t
V = USin\(\theta\) — gt
But V at maximum height = O
.-. O = Usin\(\theta\) - gt
Hence, t = \(\frac{U sin \theta}{g}\)
But time for a projectile to reach maximum he is equal to time to return to projection plane
Time of flight T = \(\frac{2Usin \theta}{g}\)
Vertical component velocity Vy = Usin\(\theta\)
Time to reach maximium height = t
V = USin\(\theta\) — gt
But V at maximum height = O
.-. O = Usin\(\theta\) - gt
Hence, t = \(\frac{U sin \theta}{g}\)
But time for a projectile to reach maximum he is equal to time to return to projection plane
Time of flight T = \(\frac{2Usin \theta}{g}\)