Find the time of flight of a projectile throw | vedclass

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Question

(D) For a projectile thrown horizontally with speed $u$ from an inclined plane making an angle $	heta$ with the horizontal,the effective acceleration

Vedclass · Accepted Answer

$10 \ s$

Vedclass · Answer

(D) For a projectile thrown horizontally with speed $u$ from an inclined plane making an angle $	heta$ with the horizontal,the effective acceleration perpendicular to the plane is $g \cos 	heta$ and the initial velocity component perpendicular to the plane is $u \sin 	heta$ (directed downwards).Using the equation of motion $s_y = u_y t + \frac{1}{2} a_y t^2$ along the axis perpendicular to the plane,where $s_y = 0$ for the total time of flight $T$:$0 = (u \sin 	heta) T + \frac{1}{2} (g \cos 	heta) T^2$Since the projectile is thrown horizontally,the initial velocity component perpendicular to the plane is $u \sin 	heta$ downwards. The acceleration component perpendicular to the plane is $g \cos 	heta$ upwards.$T = \frac{2 u \sin 	heta}{g \cos 	heta} = \frac{2 u 	an 	heta}{g}$Given $u = 50 \ m/s$,$	heta = 45^{\circ}$,and $g = 10 \ m/s^2$:$T = \frac{2 	imes 50 	imes 	an 45^{\circ}}{10} = \frac{100 	imes 1}{10} = 10 \ s$.

Find the time of flight of a projectile thrown horizontally with a speed of $50 \ m/s$ from a long inclined plane which makes an angle of $\theta = 45^{\circ}$ with the horizontal.

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