a ball is thrown from level ground at 24 m/s, 30 degree above horizontal. How much time will it take to reach its maximum height? solution

To find the time it takes for the ball to reach its maximum height, we can use the vertical component of its initial velocity. The vertical component is given by:

Vyi = V * sin(theta)

where V is the initial velocity (24 m/s) and theta is the angle above the horizontal (30 degrees).

Substituting the given values into the equation, we have:

Vyi = 24 m/s * sin(30 degrees)

Vyi = 24 m/s * 0.5

Vyi = 12 m/s

Next, we can use the equation for vertical motion to find the time it takes for the ball to reach its maximum height:

Vf = Vi + a * t

Where:
Vf is the final vertical velocity (0 m/s at maximum height)
Vi is the initial vertical velocity (Vyi)
a is the acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
t is the time.

Substituting the values into the equation:

0 m/s = 12 m/s + (-9.8 m/s^2) * t

Rearranging the equation gives us:

-12 m/s = -9.8 m/s^2 * t

Now we can solve for t:

t = -12 m/s / -9.8 m/s^2

t ≈ 1.22 seconds

Therefore, it will take approximately 1.22 seconds for the ball to reach its maximum height.