Aprojectile Is At Such An Angle That
The Vertical Component Its Volocity
Is49m\s The Horizontal Component
Of Its Volocity Is 60m\s.
A,how Long Does The Projectile
Remain In Air?
B,what Horizontal Distance Does
It Travel?
This is a relatively easy problem that is confused by two separate velocities.
When working with vertical velocity, Vy, you ignore what the horizontal velocity, Vx, is doing, and visa versa.
For A you can use this equation to solve for time until velocity is 0, which would be when the object is at its apex, and then multiply it by 2 to get the entire time it is in the air.
t=(0-Vi)/a
t=(0-49)/-9.8 <--- notice we only use the vertical component of the velocity for this, the horizontal does not matter. notice also that the acceleration from gravity is negative because it is in the opposite direction of our positive vertical velocity.
t=-49/-9.8=5. Time until velocity=0 at the top of the parabola is 5 seconds. Since no other force is acting on it you can multiply this by 2 to equal the entire time the projectile in flight, 10 seconds.
Now we have the entire time the object is moving vertically, this will also be the total time the object is moving horizontally. We have the speed that the object is moving horizontally, Vx=60m/s. Therefore in 10 seconds of travel the object will travel 600 meters horizontally.
Xo = 60 m/s.
Yo = 49 m/s.
Vo = 60 + 49i = 77.5m/s[39.2o].
A. Y = Yo + g*Tr = 0.
49 + (-9.8)Tr = 0,
Tr = 5 s. = Rise time.
Tf = Tr = 5 s. = Fall time.
Tr+Tf = 5+5 = 10 s. = Time in air.
B. d = Xo*(Tr+Tf) = 60*10 = 600 m.
To answer these questions, we need to understand the motion of a projectile. The vertical and horizontal components of velocity and the initial angle of projection play a crucial role in determining the time of flight and horizontal distance traveled. Let's break it down step by step:
Step 1: Find the time of flight (A)
The time of flight is the total time the projectile remains in the air. To calculate it, we need to find the vertical component of the initial velocity (Vv) and use it to determine how long it takes for the projectile to reach the ground.
Given:
Vertical component of velocity (Vv) = 49 m/s
Using the formula:
Vv = V * sin(θ)
Where:
Vv is the vertical component of velocity,
V is the initial velocity, and
θ is the angle of projection.
Rearranging the formula:
V = Vv / sin(θ)
Since we are given the horizontal component of velocity (Vh) and the angle of projection is not provided, we'll calculate it using the formula:
Vh = V * cos(θ)
Using the given values:
Vh = 60 m/s
Rearranging the formula:
V = Vh / cos(θ)
Now, we can find the angle of projection (θ). Rearranging either of the above formulas, we get:
θ = arctan(Vv / Vh)
Calculating the angle:
θ = arctan(49 / 60) ≈ 43.7°
Finally, to calculate the time of flight (t):
t = (2 * Vv) / g
Where:
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the known values:
t = (2 * 49) / 9.8
t = 9.8 seconds
Answer: The projectile remains in the air for approximately 9.8 seconds.
Step 2: Find the horizontal distance traveled (B)
The horizontal distance traveled by the projectile is determined by the horizontal component of velocity (Vh) and the time of flight (t).
To calculate the horizontal distance:
d = Vh * t
Substituting the known values:
d = 60 * 9.8
d = 588 meters
Answer: The projectile travels approximately 588 meters horizontally.
By following these steps, you can find the time of flight and horizontal distance traveled for a projectile launched at a given angle with known velocity components.