Answer:
Straight line parallel to time axis.
Explanation:
The slope of the position time graph gives the velocity.
As the man is still, that means the velocity is zero. So, the slope of the graph is zero. It is a straight line parallel to time axis.
A bullet is shot from shoulder height (1.50m)at an angle 60.0°above the horizontalwith an initial speed of 400m/s.Neglect air re
sistance.a. How much time elapses before the bullet hits the ground?b. How far does the bullet travel horizontally?c. What is the bullet’simpact anglewhen landing on the ground?
2 answers:
Answer:
a) t₁ = 70,717 s , b) x = 14143.4 m , c) θ = -60.02º
Explanation:
We will work on this problem using kinematics,
a) time to reach the ground, let's look for the initial speed
= v₀ sin.60
v₀ₓ = v₀ cos 60
v_{oy} = 400 sin. 60 = 346.41 m / s
v₀ₓ = 400 cos 60 = 200 m / s
Let's use the projectile launch equation
y = y₀ + v_{oy} t - ½ g t²
Upon reaching the ground y = 0
0 = 1.5+ 346.41 t - ½ 9.8 t²
4.9 t² - 346.41 t - 1.5 = 0
t² - 70.696 t - 0.306 = 0
We solve the second degree equation
t = [70.696 ±√ (70.696² + 4 0.306)] / 2
t₁ = 70,717 s
t₂ = -0.0085 s
We take the positive time
b) let's use the equation
x = v₀ₓ t
x = 200 70,717
x = 14143.4 m
c) to find the angle let's look for the velocity at the point of impact, the velocity on the x-axis is constant
vₓ = v₀ₓ = 200 m / s
The speed on the y axis
= v_{oy} - g t
v_{y} = 346.41 - 9.8 70.717
v_{y} = -346.62 m / s
Let's use trigonometry
tan θ = v_{y} / vₓ
θ = tan⁻¹ v_{y} / vₓ
θ = tan⁻¹ (-346.62 / 200)
θ = -60.02º
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