The equation above expreser the approximate height \( h=-4.5 p^{2}+25 t \) meters, of a ball launched vertically upward focoonds after it is initial velocity of 25 meters per the ground with an approximately how many seconds will the ball hit the ground? A) 3.5 B) 4.0 C) 4.5 D) 5.0

To find when the ball hits the ground, set \( h = 0 \): \[ 0 = -4.5p^2 + 25p \] Rearranging gives: \[ 4.5p^2 - 25p = 0 \] Factoring out \( p \): \[ p(4.5p - 25) = 0 \] This gives us two solutions: \( p = 0 \) (the launch moment) and \( 4.5p - 25 = 0 \), leading to \( p = \frac{25}{4.5} \approx 5.56 \) seconds. Since 5.56 isn’t an option, check for precision. To determine how many solutions exist, recognize that the other valid option is where p values to a maximum. When plugged back in, the height approaches zero closest with 4.5 seconds. The approximate answer closest to a valid solution for time would be **D) 5.0 seconds**. --- While ball trajectories can be fascinating in math, historically, they have been tied to discussions by great physicists like Galileo, who studied how objects fall under gravity. His experiments laid the groundwork for understanding motion, impacting everything from sports to space exploration! When solving quadratic equations, a common pitfall involves misspeaking or misapplying the quadratic formula. A key tip is to double-check your initial setup: ensure your equation is correctly rearranged with all terms present. This will save you from chasing after wrong answers and wasted effort!