In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0800 s, during which time it experiences an acceleration of 229 m/s2. The ball is launched at an angle of 52.3 ° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.

Question

anonymous · Answer

$$v = u + at ,where u = 0, a = 223 m/s ^{2} ,t = 0.0800 seconds$$
$$v = 0.08\times223 = 17.84 m/s$$
$$Angle of projection, \theta = 52.3( degrees)$$
$$v  \times \cos \theta = Horizontal$$
$$v \times \sin \theta = Vertical$$
Thus Horizontal Component of velocity = 17.84 * cos(52.3) = 10.909 m/s
Thus Vertical Component of velocity = 17.84 * sin(52.3) = 14.115 m/s