Question

In: Physics

A projectile is launched at an angle of 30 degrees above the horizontal at a speed...

A projectile is launched at an angle of 30 degrees above the horizontal at a speed of 300 m/s. In what direction is it traveling after 3 s? In other words, what is the direction of its velocity vector? (Use |g| = 10 m/s2.)

Expert Solution

v_o = initial velocity = 300 m/s

= angle of launch = 30

Consider the motion along the vertical direction or Y-direction

V_oy = initial velocity In Y-direction = v_o Sin = (300) Sin30 = 150 m/s

a = acceleration = - 10

t = time of travel = 3 sec

V_fy = final velocity after time "t" = ?

using the equation

V_fy = V_oy + at

V_fy = 150 + (- 10) (3)

V_fy = 120 m/s

Consider the motion along the horizontal direction or X-direction

V_ox = initial velocity In X-direction = v_o Cos = (300) Cos30 = 259.8 m/s

a = acceleration = 0

t = time of travel = 3 sec

V_fx = final velocity after time "t" = ?

using the equation

V_fx = V_ox + at

V_fx = 259.8 + (0) (3)

V_fx = 259.8 m/s

= angle after time "t"

= tan^-1(V_fy /V_fx ) = tan^-1(120/259.8 ) = 24.8 deg

genius_generous answered 6 months ago

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