A body is projected at a velocity of 200m/s and at an angle of 30 degree to the horizontal. Find a. Time of flight. b. Range. c. Maximum height.
Vi = initial vertical velocity = 200 m/s * sin(30º)
h = horizontal velocity = 200 m/s * cos(30º)
a. T = time of flight = 2 Vi / g
b. R = range = h * T
c. H = max height = (Vi ^ 2) / (2 * g)
To find the time of flight, range, and maximum height of a projectile, we can use the equations of motion for projectile motion.
a. Time of Flight:
The time of flight is the total time the projectile is in the air. We can find it using the equation:
Time of flight = (2 * initial velocity * sin(angle))/acceleration due to gravity
Here, the initial velocity is 200 m/s and the angle is 30 degrees. The acceleration due to gravity is approximately 9.8 m/s².
Plugging in the values, we get:
Time of flight = (2 * 200 * sin(30))/9.8
= (400 * 0.5)/9.8
= 20.41 seconds (approximately)
Therefore, the time of flight is approximately 20.41 seconds.
b. Range:
The range is the horizontal distance covered by the projectile. We can calculate it using the equation:
Range = (initial velocity² * sin(2 * angle))/acceleration due to gravity
Using the values we already have:
Range = (200² * sin(2 * 30))/9.8
= (40000 * sin(60))/9.8
= (40000 * √3/2)/9.8
= 6940.72 meters (approximately)
Therefore, the range is approximately 6940.72 meters.
c. Maximum Height:
The maximum height reached by the projectile can be found using the equation:
Maximum height = (initial velocity² * sin²(angle))/(2 * acceleration due to gravity)
Substituting the known values:
Maximum height = (200² * sin²(30))/(2 * 9.8)
= (40000 * 0.25)/(2 * 9.8)
= 510.2 meters (approximately)
Therefore, the maximum height reached by the projectile is approximately 510.2 meters.