uestionsPhysics
A projectile is fired at an angle of 60degree with the horizontal and with the initial velocity 80m/s calculate (a) the time of flight(b)time to reach maximum height(c)it range(d)the velocity of projection 2seconds after been fired
To answer these questions, we will use the basic equations of projectile motion. Let's break down each question and solve them step by step.
(a) The time of flight:
To calculate the time of flight, we need to find the total time it takes for the projectile to return to the same horizontal level from where it was launched. The equation to find the time of flight is given by:
Time of flight (T) = (2 * initial velocity * sin(angle))/acceleration due to gravity
Using the given values:
Initial velocity = 80 m/s
Angle = 60 degrees
Acceleration due to gravity (g) = 9.8 m/s^2
T = (2 * 80 * sin(60))/9.8
Now, calculate the value of sin(60) which is √3/2:
T = (2 * 80 * √3/2)/9.8
T = 160√3/9.8
T ≈ 9.24 seconds
Therefore, the time of flight is approximately 9.24 seconds.
(b) The time to reach maximum height:
The projectile reaches its maximum height when its vertical velocity becomes zero. To find the time taken to reach maximum height, we can use the following equation:
Time to reach maximum height (t) = initial velocity * sin(angle) / acceleration due to gravity
Using the given values:
Initial velocity = 80 m/s
Angle = 60 degrees
Acceleration due to gravity (g) = 9.8 m/s^2
t = 80 * sin(60) / 9.8
Now, calculate the value of sin(60) which is √3/2:
t = 80 * √3/2 / 9.8
t = 40√3/9.8
t ≈ 2.29 seconds
Therefore, the time taken to reach maximum height is approximately 2.29 seconds.
(c) The range:
The range is the horizontal distance covered by the projectile. To calculate it, we need to find the horizontal velocity and then multiply it by the time of flight. The equation to find the horizontal velocity (Vx) is given by:
Horizontal velocity (Vx) = initial velocity * cos(angle)
Using the given values:
Initial velocity = 80 m/s
Angle = 60 degrees
Vx = 80 * cos(60)
Now, calculate the value of cos(60) which is 1/2:
Vx = 80 * 1/2
Vx = 40 m/s
Now, we can calculate the range using the formula:
Range = Horizontal velocity * time of flight
Range = 40 * 9.24
Range ≈ 369.6 meters
Therefore, the range of the projectile is approximately 369.6 meters.
(d) The velocity of projection 2 seconds after being fired:
To find the velocity of projection after 2 seconds, we need to calculate the horizontal and vertical components of the velocity separately.
Horizontal velocity (Vx) remains constant throughout the motion, which we calculated as 40 m/s.
Vertical velocity (Vy) can be calculated using the equation:
Vertical velocity (Vy) = initial velocity * sin(angle) - g * time
Using the given values:
Initial velocity = 80 m/s
Angle = 60 degrees
Acceleration due to gravity (g) = 9.8 m/s^2
Time = 2 seconds
Vy = 80 * sin(60) - 9.8 * 2
Now, calculate the value of sin(60) which is √3/2:
Vy = 80 * √3/2 - 9.8 * 2
Vy = 40√3 - 19.6
Vy ≈ 37.60 m/s
Therefore, the velocity of projection 2 seconds after being fired is approximately 40 m/s (horizontal component) and 37.60 m/s (vertical component).
Please note that these calculations assume ideal conditions (no air resistance) and are simplified for the sake of explanation. In real-world scenarios, the projectile's motion may be affected by various factors.