If a ball has an initial speed of 26.9 m/s and is launched at an angle of 63.2 degrees above the horizontal, what is the tallest obstacle that the ball can clear?
Vo = 26.9m/s[63.2o]
Xo = 26.9*cos63.2 = 12.13 m/s.
Yo = 26.9*sin63.2 = 24.01 m/s.
Y^2 = Yo^2 + 2g*h
h = (Y^2-Yo^2)/2g = (0-24.01^2)/-19.6 =
29.41 m.
To find the tallest obstacle that the ball can clear, we need to determine the maximum height achieved by the ball's projectile path.
First, let's break down the initial velocity of the ball. The initial speed of 26.9 m/s can be split into two components: the vertical component (upward) and the horizontal component.
The vertical component can be calculated using the formula V_y = V * sin(theta), where V_y is the vertical component of velocity, V represents the initial speed, and theta is the launch angle.
So, V_y = 26.9 m/s * sin(63.2°).
Next, to find the time it takes for the ball to reach its peak height, we can use the equation V_y = V_initial_y + (acceleration * time), where the initial vertical velocity is V_initial_y, acceleration is -9.8 m/s² (due to gravity), and time is the unknown.
Rearranging the equation, we get 0 = V_y + (-9.8 m/s² * t_peak). Solving for t_peak, we have t_peak = -V_y / (-9.8 m/s²).
Now, we can find the peak height by using the formula h_peak = V_initial_y * t_peak + (1/2) * acceleration * t_peak^2.
Substituting the values, h_peak = V_y * t_peak + (1/2) * (-9.8 m/s²) * (t_peak^2).
Finally, the tallest obstacle that the ball can clear is equal to the peak height achieved. Solve the above equation to determine the value of h_peak.
Note: It's important to convert the launch angle from degrees to radians when using trigonometric functions.