Keisha throws a baseball with an initial velocity of 30 m/s at 30.0° above the horizontal. If air resistance is negligible, how long does it take for the baseball to reach the maximum height of its trajectory?

Assume that the acceleration due to gravity is given by: g = 10 m/s^2
sin(30.0°) = 1/2
cos(30.0°) = √(3) / 2

v = 30*sin30° - 9.8t

it reaches max height when v=0.

1.5s

To find out how long it takes for the baseball to reach the maximum height of its trajectory, we can use the equation of motion for vertical motion. The equation is:

y = yo + voy*t - (1/2)*g*t^2

Where:
y = vertical displacement (in this case, the maximum height)
yo = initial vertical position (which is 0 in this case)
voy = initial vertical velocity (in this case, the vertical component of the initial velocity)
g = acceleration due to gravity (given as 10 m/s^2)
t = time taken

Since the baseball is thrown at an angle of 30.0° above the horizontal, we need to find the vertical component of the initial velocity (voy). We can do this by multiplying the initial velocity (30 m/s) by the sine of the angle:

voy = 30 m/s * sin(30.0°)
voy = 30 m/s * (1/2)
voy = 15 m/s

Now we can substitute the values into the equation and solve for time (t):

y = 0 + 15 m/s*t - (1/2)*10 m/s^2*t^2

Since we want to find the time it takes to reach the maximum height, the vertical displacement (y) is equal to 0 at that point. Thus, we can simplify the equation to:

0 = 15 m/s*t - 5 m/s^2*t^2

This is a quadratic equation, so we can solve it by factoring or applying the quadratic formula. For simplicity, let's factor it:

0 = t(15 - 5t)

Now we have two possible solutions:

t = 0 or t = 15/5

Since we are looking for the positive time it takes for the baseball to reach the maximum height, we discard the t = 0 solution. Therefore, the time it takes for the baseball to reach the maximum height is:

t = (15/5) seconds
t = 3 seconds

Therefore, it takes 3 seconds for the baseball to reach the maximum height of its trajectory.