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 \(\fra...
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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.
Answer Details
Projectile motion and time of flight
A body is projected from ground level with speed \(u\) at an angle \(\theta\) to the horizontal. The diagram shows the parabolic path and the initial velocity resolved into its horizontal and vertical components.
Resolving the initial velocity:
Horizontal component: \(u_x = u\cos\theta\) (this stays constant because there is no horizontal force).
Vertical component: \(u_y = u\sin\theta\) (this is decelerated by gravity \(g\) acting downward).
Time to reach maximum height
Taking upward as positive, the vertical velocity at time \(t\) is
\[ v_y = u\sin\theta - g t \]At the maximum height the vertical velocity is momentarily zero, \(v_y = 0\). Let \(t\) be the time taken to reach this point:
\[ 0 = u\sin\theta - g t \]\[ t = \frac{u\sin\theta}{g} \]Time of flight
By the symmetry of the path, the time taken to fall from the maximum height back to the projection level equals the time taken to rise to it. Hence the total time of flight \(T\) is twice the time to reach maximum height:
\[ T = 2t = \frac{2u\sin\theta}{g} \]This is the required result: the time of flight \(T = \dfrac{2u\sin\theta}{g}\), where \(g\) is the acceleration of free fall.
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