An air cannon is used to shoot marshmallows down the 500 hall of the Albert Robi

Question

An air cannon is used to shoot marshmallows down the 500 hall of the Albert Robin campus building. The muzzle velocity is 100 meters per second. Neglect air resistance.
A. Given an initial launch angle of 0 degrees and an initial height of 1.5 meters, how far down the hallway does the marshmallow travel before striking the floor?
B. Show that the maximum range would be achieved by launching at 45 degrees.
C. What is the maximum range that the marshmallow can be launched?
D. What two angles can be used to launch the marshmallow ½ the maximum distance? Assume there is no problem with overhead obstacles.

Explanation / Answer

A) The initial speed of the M is u = 100 m/s The initial launch angle Theta = 0 degree Initial height y = 1.5 m The equations of motion for M are In X direction: x = u*t y =0.5*g*t^2 1.5 m = 0.5* 9.8 m/s^2 * t^2 t = 0.55 s The distance of the hallway does the marshmallow travel before striking the floor is x = 100 m/s * 0.55 s x = 55 m B) The horizontal distance travelled by a projectile projected at an angle theta above the horizontal is given by, R = u^2sin(2theta)/g As the sine function has a maximum value of +1, the maximum horizontal distance travelled by the projectile will be, for sin(2theta) = +1 2theta = 90 degree theta = 45 degree So to reach the maximum distance, the M should be projected at 45 degree angle C) The maximum range that M can launched is Rmax = u^2/g (sin(2theta)=1) = (100 m/s) ^2 / 9.8 m/s^2 = 1020.4 m D) Half the max range is Rmax/2 = 1020.4 m/2 = 510.2 m The horizontal distance travelled by the M is given by R = u^2sin(2theta)/g 510.2 m = (100 m/s)^2*sin(2theta)/9.8 m/s^2 sin(2theta) = 0.5 2(theta) = 30 degree theta = 15 degree The other angle to be projected for the same horizontal distance is Theta' = 90 - theta = 90-15 = 75 degree

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