Answer:
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!!!
