A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the maximum height attained by the ball? Round to the nearest tenth of a meter.
By definition, the height is given by: [tex]h (t) = (1/2) * (a) * (t ^ 2) + v0 * t + h0
[/tex] Where, a: acceleration v0: initial speed h0: initial height. The initial velocity in its vertical component is: [tex]v0 = 26 * sine (36)
v0 = 15.3 m / s[/tex] The initial height is: [tex]h0 = 0 m[/tex] (the ball is on the ground) The acceleration is: [tex]a = -9.8 m / s ^ 2[/tex] (acceleration of gravity) Substituting values: [tex]h (t) = (1/2) * (- 10) * (t ^ 2) + 15.3 * t + 0
[/tex] Rewriting: [tex]h (t) = -4.9 * t ^ 2 + 15.3 * t
[/tex] The time in which the maximum height occurs is obtained by deriving: [tex] h '(t) = -9.8 * t + 15.3
[/tex] We set zero and let's time: [tex]-9.8 * t + 15.3 = 0
t = 15.3 / 9.8
t = 1.56[/tex] We evaluate the time obtained in the equation of the height, to obtain the maximum height: [tex]h (1.56) = -4.9 * (1.56) ^ 2 + 15.3 * (1.56)
h (1.56) = 11.9 m[/tex] Answer: The maximum height attained by the ball is: h (1.56) = 11.9 m
