The distance between two cities is 480 km. What is the minimum time taken by a projectile that can be projected at 300 km/h to travel this distance.
The distance between the cities is 480 km. The minimum time taken by a missile launched from one to the other has to be determined if it is known that the missile can travel at a maximum velocity of 300 km/h.
Assume the missile is launched at an angle A to the horizontal. The speed of the missile in m/s is 250/3 = 83.33 m/s. There is an acceleration due to the force of gravity acting on the missile in the downward direction of 9.8 m/s^2.
The speed of the missile can be divided two a vertical component of `(250/3)*sin A` and a horizontal component of `(250/3)*cos A` . In the time that it the missile to travel 480000 m which is` 480000/((250/3)*cos A)` the missile moves upwards till its speed is 0 and then descends till its final speed is `(250/3)*sin A` again.
Use the formula v = u - g*t where v = 0, u = `(250/3)*sin A` and g = 9.8
=> t = `((250/3)*sin A)/9.8`
This is the time taken to reach the highest point, the total time taken to go up and come down is `((2*250/3)*sin A)/9.8` In this time the horizontal distance traveled is `480000 = ((250/3)*sin A)/4.9*(250/3)*cos A`
=> `480000 = ((250/3)^2*sin 2A)/9.8`
=> `sin 2A = 677.376`
This is not possible as sin 2A lies in the range [-1, 1]
At the maximum velocity of the missile it is not possible for it to travel the distance of 480 km between the two cities.