A player kicks a footbal with a velocity of 27 m/s at an angle of 31° with the ground. How high does the ball travel?
Vo = 27m/s @ 31deg.
Vo(ver) = 27sin31 = 13.91m/s.
Vf = Vo + gt = 0 @ max. height.
13.91 - 9.8t = 0,
9.8t = 13.91,
t = t = 1.42s = time to reach max. ht.
hmax=13.91*1.42 + 0.5*(-9.8*(1.42)^2,
hmax = 19.75 - 9.85 = 9.9m.
To determine the maximum height the ball reaches, we need to analyze the vertical motion of the ball.
First, we can break down the initial velocity into its vertical and horizontal components. The horizontal component of the velocity doesn't affect the vertical motion, so we will focus on the vertical component.
The vertical component of the initial velocity can be calculated using the equation:
Vertical velocity (V_y) = initial velocity (V) * sin(angle)
Plugging in the values:
V_y = 27 m/s * sin(31°)
Next, we can use the kinematic equation to calculate the height:
Vertical displacement (Δy) = (V_y^2) / (2 * acceleration due to gravity)
The acceleration due to gravity is approximately 9.8 m/s^2.
Plugging in the values:
Δy = (V_y^2) / (2 * 9.8 m/s^2)
Now, we can calculate the height by substituting the value of V_y from the previous calculation:
Δy = (27 m/s * sin(31°))^2 / (2 * 9.8 m/s^2)
Evaluating this equation will give you the height the ball reaches.