A 5.1 oz baseball is hit from 2 ft above the ground. It has an initial velocity of 165 ft/s at an angle 40 degrees above horizontal. Find the Kinetic Energy of the ball at its maximum height.
Work:
At max height, the vertical velocity is zero. the horizontal velocity is constant since there is no horizontal acceleration [V=165*cos(40)]. Kinetic Energy (KE)=(1/2)mv^2 {(1/2)(5.1/16)(165*cos(40))}=2546.2 lb ft^2/s^2
I was told that is not the correct answer.
To find the kinetic energy of the ball at its maximum height, we need to calculate the velocity of the ball at that point.
We know that at maximum height, the vertical velocity is zero. This means that the vertical component of the initial velocity becomes zero.
The vertical component of the initial velocity can be calculated using the formula: V_y = V_initial * sin(angle)
Given that the initial velocity is 165 ft/s and the angle is 40 degrees, we can calculate the vertical component of the initial velocity as:
V_y = 165 ft/s * sin(40 degrees)
V_y = 107.05 ft/s
Now, to find the time taken for the ball to reach its maximum height, we can use the equation: V_final = V_initial + a * t, where V_final = 0, V_initial = V_y, and a is the acceleration due to gravity (-32.2 ft/s^2).
0 = 107.05 ft/s + (-32.2 ft/s^2) * t
Solving for t, we get:
t = -107.05 ft/s / (-32.2 ft/s^2)
t = 3.325 seconds (approximately)
Now, we can use the time taken to find the horizontal distance traveled by the ball. The horizontal distance can be calculated using the equation: d = V_x * t, where V_x is the horizontal component of the initial velocity.
V_x = V_initial * cos(angle)
V_x = 165 ft/s * cos(40 degrees)
V_x = 126.39 ft/s (approximately)
d = 126.39 ft/s * 3.325 seconds
d = 420.2 feet (approximately)
Therefore, the horizontal distance traveled by the ball when it reaches its maximum height is approximately 420.2 feet.
Since the ball reaches its maximum height, it means that the total energy (including kinetic energy) is conserved. This means that the kinetic energy of the ball at its maximum height is the same as the kinetic energy of the ball at the initial point.
Using the formula for kinetic energy: KE = (1/2) * mass * velocity^2
Given that the mass of the ball is 5.1 oz, which is equivalent to 5.1/16 lb, and the initial velocity is 165 ft/s, the kinetic energy at the maximum height is:
KE = (1/2) * (5.1/16) lb * (165 ft/s)^2
KE = 848.64 lb*ft^2/s^2 (approximately)
Hence, the correct answer for the kinetic energy of the ball at its maximum height is approximately 848.64 lb*ft^2/s^2.