A batted baseball is hit with a velocity of 47.3 m/s, starting from an initial height of 8 m. Find how high the ball travels in two cases:
(a) a ball hit directly upward and
(b) a ball hit at an angle of 70° with respect to the horizontal
Also find how long the ball stays in the air in each case.
case a
case b
See your 9-12-11, 1:10pm post for solution.
To calculate the height the ball travels and the time it stays in the air in each case, we can use the kinematic equations of motion.
(a) In case a, where the ball is hit directly upward, the initial velocity in the vertical direction is given as 47.3 m/s in the positive direction (upwards). The acceleration due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction to the initial velocity), and the initial height is 8 m.
To find the maximum height reached by the ball, we need to determine the time it takes to reach its peak height. We can use the kinematic equation:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time taken.
At the peak of its trajectory, the final velocity is zero (vf = 0). Plugging in the values:
0 = 47.3 - 9.8t
Solving for t, we get t = 4.84 seconds.
To find the maximum height, we can use another kinematic equation:
Δy = vi*t + (1/2)at^2
where Δy is the change in height.
Δy = 47.3(4.84) + (1/2)(-9.8)(4.84)^2
Δy = 114.4 meters
Therefore, the ball reaches a maximum height of 114.4 meters and stays in the air for 4.84 seconds.
(b) In case b, where the ball is hit at an angle of 70° with respect to the horizontal, we need to consider both the horizontal and vertical components of velocity. The initial velocity of 47.3 m/s can be split into a horizontal component and a vertical component.
Horizontal component: vix = vicosθ = 47.3*cos(70°) = 16.99 m/s
Vertical component: viy = visinθ = 47.3*sin(70°) = 44.63 m/s
For the horizontal motion, there is no acceleration, so the time spent in the air can be determined using the horizontal component of velocity:
t = Δx / vix
Since the ball is hit at an angle, the horizontal distance covered, Δx, is the same as the horizontal component of displacement. The ball will return to the same horizontal position it started from, so Δx = 0. Therefore, the time spent in the air for case b is 0 seconds.
For the vertical motion, we can use the equation:
Δy = viy*t + (1/2)ay*t^2
where Δy is the vertical displacement, viy is the initial vertical velocity, ay is the acceleration in the vertical direction (due to gravity), and t is the time.
Substituting the values:
Δy = 44.63*t + (1/2)(-9.8)t^2
To find the time it takes for the ball to reach the ground, we set Δy = 0:
0 = 44.63*t + (1/2)(-9.8)t^2
This is a quadratic equation. Solving for t, we find two possible solutions: t = 0 seconds (which we already know) and t = 9.18 seconds.
Therefore, in case b, the ball reaches a maximum height determined by the vertical component of velocity and stays in the air for a total time of 9.18 seconds.