(a) The projectile remains in the air for 10 seconds
(b) It travels a horizontal distance of 600 meters
Step-by-step explanation:
A projectile is fired at such an angle that
1. The vertical component of its velocity is 49 m/sec
2. 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 acceleration
of 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 meters
Learn more:
You can learn more about the component of velocity in brainly.com/question/4464845
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