A bullet is fired with a certain velocity at an angle above the horizontal at a location where g= 10.0 m/s^2. The initial x any components of its velocity are 86.6 m/s and 50 m/s respectively. How long does it take before the bullet hits the ground?
v₀=sqrt{v₀² (x) +v₀²(y)}=
=sqrt{86.6²+50²} =
=100 m/s.
tan α= v₀(y)/v₀ (x)=50/86.6
α= tan ⁻¹(50/86.6) =30°
t= 2vₒ•sinα/g
To find the time it takes for the bullet to hit the ground, we need to determine the vertical components of its motion.
Let's consider the given information:
Initial velocity in the x-direction (horizontal direction): Vx = 86.6 m/s
Initial velocity in the y-direction (vertical direction): Vy = 50 m/s
Acceleration due to gravity: g = 10.0 m/s^2
First, we can determine the time it takes for the bullet to reach its maximum height (the highest point in its trajectory). At this point, the vertical component of its velocity becomes zero.
Using the equation: Vy = Voy - gt
0 = 50 - 10t
t = 50/10
t = 5 seconds
Therefore, it takes 5 seconds for the bullet to reach its maximum height.
Now, we need to find the total time of flight. Since the time of flight is symmetric, meaning the same amount of time is required to reach the highest point as it does to reach the ground, the total time of flight is twice the time taken to reach the maximum height.
Total time of flight = 2 * 5 seconds
Total time of flight = 10 seconds
Hence, it takes 10 seconds for the bullet to hit the ground.
To determine the time it takes for the bullet to hit the ground, we can analyze the vertical motion of the bullet. We'll assume there is no air resistance.
Let's break the initial velocity of the bullet into its vertical and horizontal components:
Vertical component (Vy): 50 m/s
Horizontal component (Vx): 86.6 m/s
The vertical motion of the bullet can be described using the equation:
y = y0 + Vyt - (1/2)gt^2,
where:
y is the vertical displacement (height) of the bullet,
y0 is the initial height (which we'll assume to be at ground level, y0 = 0),
Vy is the vertical component of the initial velocity (50 m/s),
g is the acceleration due to gravity (10.0 m/s^2), and
t is the time.
Since the bullet will hit the ground when y = 0, we can set up the following equation:
0 = 0 + (50)t - (1/2)(10)t^2.
Simplifying, we have:
5t^2 - 50t = 0.
Factoring out t, we get:
t(5t - 50) = 0.
This equation has two solutions: t = 0 (which is not meaningful in this context) and 5t - 50 = 0.
Solving 5t - 50 = 0, we find:
5t = 50
t = 50/5
t = 10 seconds.
Therefore, it will take the bullet 10 seconds to hit the ground.