Physics MCAT 69 Help?
A particle has an initial horizontal velocity of 1.8 m/s and an initial upward velocity of 4.7 m/s. It is then given a horizontal accelera- tion of 2.3 m/s2 and a downward acceleration of 1.6 m/s2.
What is its speed after 1.9 s? Answer in units of m/s
What is the direction of its velocity at this time with respect to the horizontal? Answer between −180◦ and +180◦.
Answer in units of ◦
- helloLv 69 years agoFavorite Answer
Vf = Vi + at
Vfx = 1.8 + 2.3 x 1.9 work it out
Vfy = 4.7 + (-1.6 x 1.9) work it out.
V^2 = Vfx^2 +Vfy^2 solve for V
at theta = arctan(Vfy/Vfx) use degree mode.