Answer to Question #234970 in Physics for Potchie

Question

A football player kicks a ball into the air with an initial velocity of 30 m/s at an angle of 60 degree relative to the ground.

A.What is the position of the football after 1 second along the x and y axis?
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B.What is the velocity after 1 s?
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C.What is its total time of flight?

Answer 1

(A) We can find the position of the football after 1 second along the x and y axis as follows:

$$x(t)=v_0tcos\theta,$$ $$x(t=1\ s)=30\ \dfrac{m}{s}\cdot1\ s\cdot cos60^{\circ}=15\ m.$$ $$y(t)=v_0tsin\theta-\dfrac{1}{2}gt^2,$$ $$y(t=1\ s)=30\ \dfrac{m}{s}\cdot1\ s\cdot sin60^{\circ}-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(1\ s)^2=21.1\ m.$$

(B)

$$v_x=v_0cos\theta=30\ \dfrac{m}{s}\cdot cos60^{\circ}=15\ \dfrac{m}{s},$$ $$v_y=v_0sin\theta-gt,$$ $$v_y=30\ \dfrac{m}{s}\cdot sin60^{\circ}-9.8\ \dfrac{m}{s^2}\cdot1\ s=16.18\ \dfrac{m}{s}.$$

Finally, we can find the velocity after 1 s from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(15\ \dfrac{m}{s})^2+(16.18\ \dfrac{m}{s})^2}=22\ \dfrac{m}{s}.$$

(C) Let's first find the time that the football takes to reach its maximum height:

$$v_y=v_{0y}-gt,$$ $$0=v_0sin\theta-gt,$$ $$t=\dfrac{v_0sin\theta}{g}=\dfrac{30\ \dfrac{m}{s}\cdot sin60^{\circ}}{9.8\ \dfrac{m}{s^2}}=2.65\ s.$$

Finally, we can find the total time of flight of football:

$$t_{flight}=2t=2\cdot2.65\ s=5.3\ s.$$