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 equations x = (v0 cos(α))t y = (v0 sin(α))t − 1 2 gt2 where 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 = 500 m/s. When will the bullet hit the ground?

a) The bullet hits the ground 51.02 s after it was fired.

b) 22,092.3 units

c) 3188.8 units

Explanation:

a) Assuming that the level the bullet was fired from is the ground level.

The bullet hits the ground when y = 0

y = (v₀ sin(α))t − (1/2) gt²

v₀ = 500 m/s

α = 30°

y = 0

0 = (500 sin 30) t - 0.5(9.8)t²

4.9t² - 250t = 0

t(4.9t - 250) = 0

t = 0 s or (4.9t - 250) = 0

The t = 0 s indicates that the bullet was indeed fired from the ground level.

The time it eventually hits the ground back

4.9t = 250

t = 51.02 s

The bullet hits the ground 51.02 s after it was fired.

b) The distance from the firing point that the bullet lands.

x = (v₀ cos(α))t

At this horizontal distance, t = 51.02 s

Substituting the parameters

x = (500 cos 30°) × 51.02

x = 22,092.3 units

c) Maximum height attained by the bullet

Maximum height is given by

H = (u² sin² α)/2g

H = (500² sin² 30°)/(2×9.8)

H = 3188.8 units

Hope this Helps!!!

this a false

explanation:

need to see circuit to get time constant