If a projectile is fired with an initial velocity of v0 meters per second at an angle α above the horizontal and air resistance is assumed to be negligible, then its position after t seconds is given by the parametric equationsx = (v0 cos α) t y = (v0 sin α) t - 1/2gt^2where g is the acceleration due to gravity (9.8 m/s2). (Round your answers to the nearest whole number.) a. If a gun is fired with α = 30° and v0 = 600 m/s. When will the bullet hit the ground? b. How far from the gun will it hit the ground? c. What is the maximum height reached by the bullet? d. Find the equation of the parabolic path by eliminating the parameter.

Question

Physics

Isaiah Finley

24 February, 21:24

If a projectile is fired with an initial velocity of v0 meters per second at an angle α above the horizontal and air resistance is assumed to be negligible, then its position after t seconds is given by the parametric equationsx = (v0 cos α) t y = (v0 sin α) t - 1/2gt^2where g is the acceleration due to gravity (9.8 m/s2). (Round your answers to the nearest whole number.) a. If a gun is fired with α = 30° and v0 = 600 m/s. When will the bullet hit the ground? b. How far from the gun will it hit the ground? c. What is the maximum height reached by the bullet? d. Find the equation of the parabolic path by eliminating the parameter.

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Answers (1)

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Fisher Ward · Answer 1

Answer: The ball will hit the ground at t = 61.22s

b) 31813.17m

c) 4591.84m

d) x = 519.62t, y=300t - 1/2 gt^2

Explanation:

From the question,

x = (Vo cosα) t

Where x = horizontal range, Vo = initial velocity α = angle of inclination, t = time taken.

The first question is when will the bullet hit the ground (that is at what time will the bullet hit the ground). Before we can get this, we need to find the value of x which is the horizontal range.

This is done below with the formula

x = Vo² sin2α/g

Where x = horizontal range, Vo = initial velocity, g = acceleration due gravity = 9.8m/s².

Thus

x = 600² x sin (60) / 9.8

x = 360000 x 0.8660 / 9.8

x = 31813.17m

To get the time, we recall that

x = (Vo cosα) t where all parameters have their normal meaning

t = x/Vo cosα

t = 31813.17 / 600 x cos 30

t = 31818.17 / 92.55

t = 61.22s.

b) how far will the gun hit the ground is simply what is the horizontal distance covered by the gun, this is referring to the horizontal range (x) which we have already gotten when solving for time (t).

Thus the gun will travel a distance of 31813.17m before it hits the ground.

c) maximum height attained formulae is given as

H = Vo² sin²α/2g

Where H = maximum height, Vo = initial velocity and g = 9.8m/s²

By slotting in, we have that

H = 600² * (sin 30) ²/2 x 9.8

H = 360000 * (0.5) ² / 19.6

H = 4591.84m

d) the equation of the parabolic path is gotten by substituting values for Vo and α into the equation below

x = (Vo cosα) t

x = (600 cos 30) t

Thus x = 519.62t

And for y = (Vo sin α) t - 1/2gt^2

y = (600 sin 30) t - 1/2gt^2

y = 300 - 1/2gt^2.

This the parametric equation are x = 519.62t, y=300t - 1/2 gt^2