Vo = 200m/s[60o],
Yo = 200*sin60 = 173.2 m/s. = Ver. component of initial velocity.
Y = Yo + g*Tr = 0,
173.2 + (-10)Tr = 0,
Tr = 17.3 s. = Rise time.
Tf = Tr = 17.3 s. = Fall time.
Tr + Tf = Time in flight.
Yo = 200*sin60 = 173.2 m/s. = Ver. component of initial velocity.
Y = Yo + g*Tr = 0,
173.2 + (-10)Tr = 0,
Tr = 17.3 s. = Rise time.
Tf = Tr = 17.3 s. = Fall time.
Tr + Tf = Time in flight.
Vertical Motion:
The vertical motion can be analyzed using the equation of motion:
y = uyt + (1/2)gt^2
where,
y = vertical displacement (0, since we are calculating the time of flight)
uy = initial vertical velocity (200m/s * sin(60°))
g = acceleration due to gravity (10m/s^2)
t = time
Using these values, we can rewrite the equation as:
0 = (200m/s * sin(60°)) * t + (1/2)(10m/s^2) * t^2
0 = (100√3)m/s * t + 5m/s^2 * t^2
Horizontal Motion:
The horizontal motion can be calculated using the equation:
x = uxt
where,
x = horizontal displacement (unknown)
ux = initial horizontal velocity (200m/s * cos(60°))
t = time
Using these values, we get:
x = (200m/s * cos(60°)) * t
x = (100m/s)t
Since the time of flight is the same for both the vertical and horizontal motions, we can equate the two equations for time:
(100√3)m/s * t + 5m/s^2 * t^2 = (100m/s) * t
Simplifying the equation, we get:
(100√3)m/s + 5m/s^2 * t = 100m/s
Solving for t, we have:
5m/s^2 * t = 100m/s - (100√3)m/s
5m/s^2 * t = (100 - 100√3)m/s
t = (100 - 100√3)m/s / 5m/s^2
t ≈ 11.54s
Thus, the total time of flight of the bullet is approximately 11.54 seconds.
First, let's consider the vertical motion of the bullet. We can use the equation of motion:
v = u + at
where:
v = final velocity (0 m/s when the bullet reaches the ground)
u = initial velocity in the vertical direction (vertical component of the initial velocity)
a = acceleration due to gravity (-10 m/s^2)
t = time of flight
Since the initial velocity is given at an angle of 60° to the horizontal, we need to find the vertical component of the initial velocity.
Vertical component (u_y) = initial velocity * sin(theta)
theta = angle of projectile (60°)
initial velocity = 200 m/s
u_y = 200 m/s * sin(60°)
u_y = 200 m/s * (√3/2)
u_y = 100√3 m/s
Now, we can use the equation above to find the time taken for the bullet to reach the ground in the vertical direction:
0 = 100√3 m/s + (-10 m/s^2) * t
Solving for t:
10 m/s^2 * t = 100√3 m/s
t = 100√3 m/s / 10 m/s^2
t = 10√3 s
Now, let's consider the horizontal motion of the bullet. The time of flight in the horizontal direction is the same as the time of flight in the vertical direction.
Therefore, the total time of flight of the bullet is:
Total time of flight = t (horizontal) = t (vertical) = 10√3 s ≈ 17.32 s
So, the total time of flight of the bullet is approximately 17.32 seconds.