Question

A projectil e is projected with a speed of 3m/s at an angle 60°to horizontal.calculate a.max height b. Time of flight

Answer 1

as we know that in projectile motion Hmax=u²sin²Ф/2g
                                                                   =3²(sin 60°)²/2×10
                                                                   =9×(√3/2)²/20
                                                                   =(9×3/4)20
                                                                   =0.3375 m
                     

Answer 2

$\huge\mathfrak\red{Answer}$

MAXIMUM HEIGHT

$initial \: velocity(u) = 3m {sec}^{ -1 } \\ \alpha = 60 \\ \\ maximum \: height = \frac{ {u}^{2} {sin}^{2} \alpha }{2g} \\ maximum \: height = \frac{ {3}^{2} {sin}^{2} 60 }{2(10) } \\ maximum \: height = \frac{9 \times \frac{3}{4} }{20} \\ \\ maximum \: height = \frac{27}{80} m$

TIME OF FLIGHT

$time \: of \: flight = \frac{2u \sin \alpha }{g} \\ \\ time \: of \: flight = \frac{2 \times 3 \times \frac{ \sqrt{3} }{2} }{10} \\ \\ time \: of \: flight = \frac{3 \sqrt{3} }{10} \\ \\ time \: of \: flight = 0.51sec$