A body is projected from ground at angle 60 degree with the horizontal. If it comes back to ground at distance 60(3)^1/2 meters from starting point then what is the time of flight?

1 Answer
Harish Chandra Rajpoot
Jul 11, 2018

#6\ sec#

Explanation:

Let the body be projected at an initial velocity #u\ m/s# at an angle #\theta=60^\circ# with the horizontal then the horizontal distance #R# traveled by the object

#R=(\text{horizontal velocity})\times (\text{time of flight})#

#60\sqrt3=\frac{u^2\sin2\theta}{g}#

#60g\sqrt3=u^2\sin(2\times 60^\circ)#

#60g\sqrt3=u^2\frac{\sqrt3}{2}#

#u^2=120g#

#u=\sqrt{120g}#

Now, the time of flight

#=\frac{2u\sin\theta}{g}#

#=\frac{2\sqrt{120g}\ \sin60^\circ}{g}#

#=\frac{2\sqrt{120}\ \sqrt3/2}{\sqrtg}#

#=\frac{6\sqrt10}{\sqrt10}\quad (\text{taking}\ g=10\ m/s^2)#

#=6\ sec#

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