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4. When solving projectile motion problems where the projectile is launched at an angle, the first step is to find the x- and y-component of the initial velocity. Why is this helpful in solving these types of problems?

5. Candace throws a ball so that it has a horizontal velocity of 6.2 m/s, and vertical velocity
of 2.1 m/s. What is the velocity vector of this ball right as she releases it?

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Exactdotcom

Answer:

4: example: The initial velocity in the y-direction will always be 0 m/s as well. In the x component list, your initial and final velocities will be the same, since there is no new force acting on the object, meaning your acceleration will be 0 m/s^2. 7

5: 4.1 m/s.

brainsoft

Answer:

Question 4:

» Acceleration due to gravity.

• When solving projectile motion. For vertical motion, acceleration due to gravity is included whereas in horizontal motion, acceleration due to gravity is excluded.

[tex]{ \boxed{ \mathfrak{vertical \: motion : { \rm{y = u _{y} t - \frac{1}{2}gt : \{u _{y} = u \sin( \theta) \} }}}}} \\ \\ { \boxed{ \mathfrak{horizontal \: motion : { \rm{x = u _{x}t : \: \{u _{x} = u \cos( \theta) }}}}} \\ [/tex]

Question 5:

[tex]{ \tt{v _{r} {}^{2} = v _{y} {}^{2} + v _{x} {}^{2} }}[/tex]

  • v_r is the resultant velocity

[tex] { \tt{v _{r} = \sqrt{ {(2.1)}^{2} + {(6.2)}^{2} } }} \\ \\ { \tt{v _{r} = \sqrt{42.85} }} \\ \\ { \tt{v _{r} = 6.5 \: m {s}^{ - 1} }}[/tex]

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