Question

A projectile is fired at such an angle that the vertical

Answer 1

(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 = $u_{y}$ t - $\frac{1}{2}$ g t², where

$u_{y}$ 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

∵ $u_{y}$ = 49 m/s

∵ g = 9.8 m/s²

- Substitute these values in the rule above

∴ 0 = 49 t - $\frac{1}{2}$ (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 = $u_{x}$ t, where $u_{x}$ is the

   horizontal component of its velocity

∵ $u_{x}$ = 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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