Answer to Question #158394 in Mechanics | Relativity for Emmanuel

Question

A projectile is lunched with 30m/s at an angle 60 degree with the horizontal. Find the position of the projectile and the magnitude and direction of its velocity at t=2s

Answer 1

An object moving in the two-dimensional parabolic path is a projectile motion. It moves in that trajectory because it is affected only by gravitational acceleration. Projectile Motion is described by two-dimensional kinematic equations.

Position

$x_t = x_o+v_0cos(\theta_o)t$

$y_t = y_o + v_osin(\theta_o)t$

Velocity

$v_x = v_ocos(\theta_o)$

$v_y = v_osin(\theta_o)-gt$

The given in the problem:

$v_0=30\frac{m}{s} \space \space \space \space \theta_0=60^o \space\space\space\space g = 9.8 \frac{m}{s^2}$

$v_x = 30*cos60^o = 15\frac{m}{s} \space\space\space\space\space v_y =15*sin60^0-9.8*2 = -6.6\frac{m}{s}$

The negative answer indicates it is moving downward.

Using Pythagoras: $v =\sqrt{v_x^2+v_y^2} = \sqrt{15^2+(-6.6)^2} = 16.4\frac{m}{s^2}$