Question

a projectile is launched at an angle of 30 degrees and lands later at the same level. if it's initial speed is 50 m/s, solve for the time it is in the air, the maximum height it reaches, and the horizontal distance it travels.

Answer 1

`using \: the \: formula \\ t = (2u \sin( \alpha ) )/(g) where \: u = initial \: speed \: \\ \alpha = angle \: of \: projection \\ g = acceleration \: due \: to \: gravity \\ (2 * 50 * \sin(30) )/(10) \\ (100 * 0.5)/(10) = (50)/(10) = 5seconds`

Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres