Question

the horizontal and vertical displacement of a projectile at time t are x= 36t and y= 48t -4.9 t² respectively. Initial velocity of the projectile in m/s is​

Answer 1

Explanation:

$u = \sqrt{ {(u(x))}^{2} + (u(y)) {}^{2} } \\ u(x) = \frac{dx}{dt} \\ u(x) = \frac{d}{dt} 36t = 36 \\ u(y) = \frac{dy}{dt} = \frac{d}{dt} 48t - 4.9 {t}^{2} = 48 - 9.8t$