a projectile is fired at an angle of 60 degrees with the horizontal and with the initital velocity of 80 m/s. What is the time of flight? What is the max height attained and the time taken to attain it? What is the range? The velocity of projection 2 seconds after being fired?
t = 2•v(o) •sinα/g,
h= v(o)^2•(sinα) ^2/2•g,
L= v(o)^2•(sin2α)/g,
v(y) = v(oy) - g•t = v(o) •sin α - g•t,
v(x) = v(ox) = v(o) •cos α,
v = sqrt[v(x)^2 + v(y)^2]
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To solve these projectile motion problems, we can break down the initial velocity of the projectile into its horizontal and vertical components.
First, let's find the horizontal component of the initial velocity:
Horizontal component of velocity (Vx) = Initial velocity × cos(angle)
Vx = 80 m/s × cos(60°)
Vx = 80 m/s × 0.5
Vx = 40 m/s
Second, let's find the vertical component of the initial velocity:
Vertical component of velocity (Vy) = Initial velocity × sin(angle)
Vy = 80 m/s × sin(60°)
Vy = 80 m/s × 0.866
Vy = 69.28 m/s
Now we can solve the questions:
1. Time of flight:
Time of flight refers to the total time the projectile remains in the air. We can find it using the vertical component of the initial velocity:
Time of flight = (2 × Vy) / g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Time of flight = (2 × 69.28 m/s) / 9.8 m/s²
Time of flight = 14.12 seconds
2. Maximum height attained and time taken to attain it:
The maximum height is reached when the vertical component of velocity becomes zero (at the topmost point of the trajectory). We can find it using the equation:
Maximum height (H) = (Vy²) / (2 × g)
Maximum height = (69.28 m/s)² / (2 × 9.8 m/s²)
Maximum height = 240.28 meters
To find the time taken to reach the maximum height, we can use the equation:
Time taken to reach the maximum height = Vy / g
Time taken to reach the maximum height = 69.28 m/s / 9.8 m/s²
Time taken to reach the maximum height = 7.07 seconds
3. Range:
The range is the horizontal distance covered by the projectile. We can find it using the horizontal component of the initial velocity and the time of flight:
Range = Vx × Time of flight
Range = 40 m/s × 14.12 seconds
Range = 564.8 meters
4. Velocity of projection 2 seconds after being fired:
To find the velocity of projection 2 seconds after being fired, we need to consider the horizontal and vertical components separately as the horizontal velocity remains constant while the vertical velocity changes due to gravity.
The horizontal velocity remains constant at 40 m/s.
To find the vertical velocity, we can use the equation:
Vy = Initial vertical velocity - (g × time)
Vy = 69.28 m/s - (9.8 m/s² × 2 s)
Vy = 69.28 m/s - 19.6 m/s
Vy = 49.68 m/s
Therefore, the velocity of projection 2 seconds after being fired is:
Velocity = √(Vx² + Vy²)
Velocity = √((40 m/s)² + (49.68 m/s)²)
Velocity = √(1600 m²/s² + 2468.99 m²/s²)
Velocity = √4068.99 m²/s²
Velocity = 63.8 m/s