2an \( h=-4.9 p^{2}+25 t \) The equation above exprever the approximate is height \( h \), in meters, of a balf, feconds after it is an launched vertically upward from the groundrwith an initial velocity of 25 meters per second. After approximately how many seconds will the ball hit the ground? A) 3.5 B) 4.0 C) 4.5 D) 5.0

To find out when the ball will hit the ground, we need to set the height \( h \) to zero and solve for time \( t \) in the equation \( h = -4.9t^2 + 25t \). This gives us the quadratic equation: \( 0 = -4.9t^2 + 25t \). Factoring out \( t \) leads to \( t(25 - 4.9t) = 0 \). This tells us \( t = 0 \) (when launched) or \( t = \frac{25}{4.9} \approx 5.1 \) seconds. Hence, the ball will hit the ground just after 5 seconds, rounding it down suggests that option D (5.0 seconds) is the closest answer. In the context of projectile motion, it’s fascinating how the effects of gravity shape the path of an object. The negative quadratic term \( -4.9t^2 \) represents gravitational pull, while \( 25t \) indicates the initial upward velocity. So, every second that passes, gravity tugs a little more at the ball, making it slow down and eventually fall back to Earth! It’s like a cosmic dance of forces!