The quarterback of a football team releases a pass at a height of 8 feet above the playing field, and the football is caught by a receiver 38 yards directly downfield at a height of 4 feet. The pass is released at an angle of $45 ^ { \circ } \text { with the horizontal. Find the speed of the football when it is }$ released. Round your answer to three decimal places.
A)33.169 feet per second
B)59.366 feet per second
C)19.462 feet per second
D)118.732 feet per second

Question

Quizplus · Accepted Answer

We can use the kinematic equations of motion to solve the problem. Let's choose a coordinate system where up is positive, and let's use the following variables:
- y1 = 8 ft (initial height)
- y2 = 4 ft (final height)
- x = 38 yd = 114 ft (horizontal distance)
- v0 = initial velocity (what we want to find)
- g = 32.2 ft/s^2 (acceleration due to gravity)

From the given angle, we can determine that the initial velocity has both horizontal and vertical components:
v0x = v0 cos(45°)
v0y = v0 sin(45°)

At the highest point of the trajectory (when the football reaches its maximum height), the vertical velocity is zero, and we can use the following kinematic equation to find the time it takes to get there:
y2 - y1 = v0yt - 1/2 gt^2
0 - 8 = v0 sin(45°) t - 1/2 (32.2) t^2
16.1t^2 - 22.6t - 8 = 0

We can solve for t using the quadratic formula:
t = [22.6 ± sqrt(22.6^2 - 4(16.1)(-8))]/(2(16.1))
t ≈ 1.12 s

Now we can use the horizontal distance and the horizontal velocity to find v0:
x = v0xt
114 = v0 cos(45°) (1.12)
v0 = 59.366 ft/s

Therefore, the answer is (B) 59.366 feet per second.

The Quarterback of a Football Team Releases a Pass at a Height