Akicker punts a football from the very center of the field to the sideline 39 yards downfield.

what is the magnitude of the net displacement of the ball? (a, football field is 53 yards wide)

what is the angle between the direction of the net displacement of the ball and the central line of the feild.

If one were to put the y value into the equation, they could find out what x is for a certain value of y.  putting in the value of y (1), we get the equation: 1=4x-12=4xx=0.5this means that the value of x when why is 1 is x=0.5

Distance covered, s = 8ft - 4ft = 4 ftinitial velocity, u = 34 ft/sthe acrobat is moving upwards and therefore, a = -g = -9.81 m/s^2 = 32.19 ft/s^2applying equation of motion; s = ut -1/2gt^2then,4 = 34t -1/2*32.19t^24 = 34t - 16.095t^2rearranging; -16.095t^2 + 34t - 4 = 0solving the quadratic equation; t = {-34 +/- sqrt [34^2 - 4(-16.)]}/2(-16.095) = 1.06 +/- 0.93 = 0.13 or 1.99the acrobat will take 0.13 seconds to be at 8 ft above the ground. additionally, once at the maximum height, the acrobat will move down and be at the same spot (that is height) after 1.99 seconds moving down.