Q:

A projectile is fired at such an angle that the verticalcomponent of its velocity is 49 m/sec. The horizontalcomponent of its velocity is 60 m/sec. (a) How long doesthe projectile remain in the air (b) What horizontal distancedoes it travel?

Accepted Solution

A:
(a) The projectile remains in the air for 10 seconds(b) It travels a horizontal distance of 600 metersStep-by-step explanation:A projectile is fired at such an angle that1. The vertical  component of its velocity is 49 m/sec2. The horizontal  component of its velocity is 60 m/sec We need to find:(a) How long the projectile remains in the air(b) The horizontal distance it travels∵ The vertical distance y = [tex]u_{y}[/tex] t - [tex]\frac{1}{2}[/tex] g t², where[tex]u_{y}[/tex] is the vertical  component of its velocity, g is the accelerationof gravity and t is the time∵ y = 0 ⇒ it return to the same initial height∵ [tex]u_{y}[/tex] = 49 m/s∵ g = 9.8 m/s²- Substitute these values in the rule above∴ 0 = 49 t - [tex]\frac{1}{2}[/tex] (9.8) t²∴ 0 = 49 t - 4.9 t²- Take t as a common factor∴ 0 = t (49 - 4.9 t)- Equate each term by 0∴ t = 0 ⇒ at initial position∴ 49 - 4.9 t = 0- Add 4.9 t to both side∴ 49 = 4.9 t- Divide both sides by 4.9∴ t = 10 seconds(a) The projectile remains in the air for 10 seconds∵ The horizontal distance = [tex]u_{x}[/tex] t, where [tex]u_{x}[/tex] is the    horizontal component of its velocity∵ [tex]u_{x}[/tex] = 60 m/sec∵ t = 10 seconds∴ x = 60 × 10 = 600 meters(b) It travels a horizontal distance of 600 metersLearn more:You can learn more about the component of velocity in brainly.com/question/4464845#LearnwithBrainly