A plane is inclined at an angle of $3 0^{\circ}$ . A body is projected from the bottom of the inclined plane at an angle of $6 0^{\circ}$ with the ground at a speed of $20 m / s$ . Find the range and time of flight.

Question

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Answer 1

$U_{x} = 20 cos 60$
$= 10 m / s$

$U_{y} = 20 sin 60 = 103 m / s$

For time of flight
$S_{y} = U_{y} t + \frac{1}{2} a_{y} t^{2}$

$0 = 103 \times T - \frac{1}{2} \times g cos 30 \times T^{2}$

$T = \frac{2 \times 1 0 3}{g cos 3 0} = \frac{2 0 3}{1 0 \times 3} \times 2 = 4$ sec

$U_{x} = 20 cos 30 = 103 m / s$

$U_{y} = 20 sin 30 = 10 m / s$

$T = \frac{2 U _{x}}{a} = \frac{2 \times 1 0}{g cos 3 0} = \frac{2 0 \times 2}{1 0 \times 3}$

$= \frac{4}{3}$ sec

$R = U_{x} T + \frac{1}{2} a_{x} T^{2}$

$= 103 \frac{4}{3} + \frac{1}{2} . (- g sin 30) \times \frac{1 6}{3}$

$= 40 - \frac{1}{2} \times 10 \times \frac{1}{2} \times \frac{1 6}{3}$

$= 40 - \frac{4 0}{3} = \frac{8 0}{3} m$

Answer 2

U_{x} = 20 cos 60

= 10 m / s

U_{y} = 20 sin 60 = 103 m / s

For time of flight

S_{y} = U_{y} t + \frac{1}{2} a_{y} t^{2}

0 = 103 \times T - \frac{1}{2} \times g cos 30 \times T^{2}

T = \frac{2 \times 1 0 3}{g cos 3 0} = \frac{2 0 3}{1 0 \times 3} \times 2 = 4

sec

U_{x} = 20 cos 30 = 103 m / s

U_{y} = 20 sin 30 = 10 m / s

T = \frac{2 U _{x}}{a} = \frac{2 \times 1 0}{g cos 3 0} = \frac{2 0 \times 2}{1 0 \times 3}

= \frac{4}{3}

sec

R = U_{x} T + \frac{1}{2} a_{x} T^{2}

= 103 \frac{4}{3} + \frac{1}{2} . (- g sin 30) \times \frac{1 6}{3}

= 40 - \frac{1}{2} \times 10 \times \frac{1}{2} \times \frac{1 6}{3}

= 40 - \frac{4 0}{3} = \frac{8 0}{3} m