3) An artillery shell is fired at a horizontal distance d from the bottom of a cliff of height h with a muzzle velocity vo at an angle ? with respect to the horizontal, and lands a horizontal distance x from the top of the cliff, where d 3500 m, h 2400 m, and g- 9.81 m/s. The angle ? that produces the maximum horizontal distance x is given by the equation The horizontal distance x from the top of the cliff and the maximum height of the shell with respect to the ground, Ymax, are given by 2. aum m ax 2. Write a MATLAB program as follows: a) vo will go from 300 m/s to 460 m/s in steps of 40 m/s . b) For each value of vo, first call the MATLAB function fzero to calculate the angle ? that gives the maximum horizontal distance x from the top of the cliff, and then use this value f ? to calculate x and ymax . Then print Vo, ?, x and ymax . Use 70 as the initial guess for ?. The output of this program should look like this: v0-300 the ta-55.37465 x=2834.92716 ymax=3106.14570 v0-340 theta-52.40923 x-5571.79510 ymax- 3699.42792 v0=380 the tas: 5 0 . 61682 x=8583 . 64029 ????= 4 3 9 6 . 80466 v0-420 theta=49.43019 x= 1 1 8 9 5 . 7 0 913 yax-5187 . 82924 v0 460 theta 48.59606 x-15518.99724 ymax- 6067.57473