Answer:
i) The horizontal and vertical components of the softball's motion can be described by the following parametric equations:
x(t) = v0x * t
y(t) = v0y * t - 1/2 * g * t^2 + h0
where v0x and v0y are the initial horizontal and vertical velocities, respectively, g is the acceleration due to gravity (32.2 ft/s^2), h0 is the initial height (5 ft), and t is time.
We can find v0x and v0y by resolving the initial velocity vector into horizontal and vertical components:
v0x = v0 * cos(45°) = 49 ft/s * cos(45°) ≈ 34.65 ft/s
v0y = v0 * sin(45°) = 49 ft/s * sin(45°) ≈ 34.65 ft/s
Substituting these values into the parametric equations, we get:
x(t) = 34.65 * t
y(t) = 34.65 * t - 16.1 * t^2 + 5
ii) The softball will be in the air until it hits the ground, which occurs when y(t) = 0. We can solve for t using the quadratic formula:
16.1 * t^2 - 34.65 * t + 5 = 0
t ≈ 2.19 s (rounded to two decimal places)
Therefore, the softball was in the air for approximately 2.19 seconds.
iii) The horizontal distance traveled by the softball can be found by evaluating x(t) at the time when the softball hits the ground:
x(2.19) ≈ 75.8 ft (rounded to one decimal place)
Therefore, the softball traveled approximately 75.8 feet in the air.
iv) The softball reaches its maximum height when its vertical velocity is zero. We can find the time when this occurs by setting v0y - g * t = 0 and solving for t:
t = v0y / g = 34.65 / 32.2 ≈ 1.08 s (rounded to two decimal places)
Therefore, the softball reaches its maximum height after approximately 1.08 seconds.
v) The maximum height reached by the softball can be found by evaluating y(t) at the time when the softball reaches its maximum height:
y(1.08) ≈
Explanation: