To solve this problem, we can break it down into two components: the horizontal motion and the vertical motion.
First, let's analyze the vertical motion. We know that the football leaves the quarterback's hands 1.75m above the ground and the receiver wants to catch it at the same height. The vertical motion can be described using the equation:
π = π£βπ‘ + 0.5ππ‘Β²,
where π is the vertical displacement, π£β is the initial vertical velocity, π‘ is the time, and π is the acceleration due to gravity (-9.8 m/sΒ²).
In this case, π = 1.75 m, π£β = 0 m/s (since the receiver wants to catch it at the same height), and π = -9.8 m/sΒ². Substituting these values, we can solve for π‘.
1.75 = 0(π‘) + 0.5(-9.8)(π‘)Β²
1.75 = -4.9π‘Β²
π‘Β² = 1.75 / -4.9
π‘Β² β -0.357
Since time cannot be negative, we discard the negative value. Therefore, π‘ is imaginary, which means it is not possible for the receiver to catch the ball at the same height as it leaves the quarterback's hands.
Now let's move on to the horizontal motion. We know that the receiver starts 12.0m away from the quarterback along the line of flight of the ball. The horizontal motion can be described using the equation:
π = π£βπ‘,
where π is the horizontal displacement, π£β is the initial horizontal velocity, and π‘ is the time.
In this case, π = 12.0 m, and π‘ is the same value we obtained earlier (even though it was imaginary). We can solve for π£β.
12.0 = π£β(π‘)
π£β = 12.0 / π‘
Since π‘ is imaginary, there is no valid solution for π£β. Therefore, it is not possible for the receiver to have a constant velocity to catch the ball at the instant it is 1.75 m above the ground, given the initial conditions and constraints provided in the question.