Final answer:
To find the smallest positive value of time when y-coordinate is a local maximum, we use the kinematic equation y = (voy + Vy)t, setting Vy to zero, which means at that point time is t = y / voy. If given a quadratic equation such as 1² + 1 - 20 = 0, we consider only the positive time value.
Step-by-step explanation:
The question is asking us to find the smallest positive value of time (t) at which the projectile reaches a local maximum in height, or in other words, its highest point. This typically happens when the vertical component of the velocity is zero. From the information provided, we can identify that the initial vertical position, yo, is 0 and the displacement, y, is -1.0000 m (which may refer to the total displacement after some time, but isn't directly relevant for the local maximum).
To solve this problem, we can use the kinematic equation for vertical motion:
y = yo + (voy + Vy)t
Since yo is zero, it simplifies to:
y = (voy + Vy)t
Since we are interested in the time t when the vertical velocity Vy is zero (at the top of the projectile's arc), we don't need to know the value of Vy. Instead, we solve for t when Vy is zero, which gives us t = y / voy. The quadratic equation provided 1² + 1 - 20 = 0 may refer to another part of the problem involving time. However, we can use the positive solution of that quadratic equation, t = 4.0 s, as it represents the time after the ball is released at the top of the trajectory when the y-coordinate is at the maximum.