x = 39
y = 55/2 = 27.5
d = sqrt [ 39^2 + 27.5^2 ]
tan theta = y/x = 27.5 / 39
A kicker punts a football from the very center of the field to the sideline 39 yd downfield. What is the magnitude of the net displacement of the ball? (A football field is 55 yd wide.) What is the angle of the net displacement of the ball from a line down the center of the field?
3 answers
I was able to figure out "What is the magnitude of the net displacement of the ball? "
It is 48 yd
I only need help with "What is the angle of the net displacement of the ball from a line down the center of the field?"
thanks
It is 48 yd
I only need help with "What is the angle of the net displacement of the ball from a line down the center of the field?"
thanks
tan theta = y/x = 27.5 / 39
theta = tan^-1 ( 0.705) = 35.2 degrees
theta = tan^-1 ( 0.705) = 35.2 degrees