The picture above shows a football player kicking a football. this is known as two dimensional motion. in which direction does the football move? a) vertical only (y) b) horizontal only (x) c) horizontal and vertical (x, y) d) horizontal, vertical, and side to side (x, y, and z) the diagram shows the motion of a tennis ball that has just been hit with a racket (air resistance is neglected). which of these is true of the horizontal and vertical components of the ball’s velocity? a) both the horizontal and the vertical components are constant. b) both the horizontal and the vertical components are accelerated. c) the horizontal component is constant but the vertical component is accelerated. d) the horizontal component is accelerated but the vertical component is constant.

1. C)horizontal and vertical (x, y)

The picture shows the motion of a projectile, which consists of two separate motions along two different directions:

- horizontal (x): along this direction, the football has a uniform motion, with constant horizontal speed ,where is the magnitude of the initial velocity of the ball and the angle at which it has been thrown

- vertical (y): along this direction, the football has an accelerated motion, with initial vertical velocity upward and constant acceleration downward (acceleration due to gravity)

2. C)The horizontal component is constant but the vertical component is accelerated.

As described in the previous part of the exercise:

- along the horizontal direction there are no forces exerted on the ball, so it is a uniform motion, therefore the acceleration is zero and the horizontal component of the velocity is constant

- along the vertical direction there is one force acting on the ball (the force of gravity, which pushes downward), so there is an acceleration (downward) equals to and therefore the vertical component of the velocity is not constant.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft