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
