h=−4.9t2+25tThe equation above expresses the approximate height h, in meters, of a ball t seconds after it is launched vertically upward from the ground with an initial velocity of 25 meters per second. After approximately how many seconds will the ball hit the ground?Question 4 options:544.53.5

ANSWER5EXPLANATIONThe equation that expresses the approximate height h, in meters, of a ball t seconds after it is launched vertically upward from the ground is[tex]h(t) = - 4.9 {t}^{2} + 25t[/tex]To find the time when the ball hit the ground,we equate the function to zero.[tex] - 4.9 {t}^{2} + 25t = 0[/tex]Factor to obtain;[tex]t( - 4.9t + 25) = 0[/tex]Apply the zero product property to obtain,[tex]t = 0 \: or \: \: - 4.9t + 25 = 0[/tex][tex]t = 0 \: \: or \: \: t = \frac{ - 25}{ - 4.9} [/tex]t=0 or t=5.1 to the nearest tenth.Therefore the ball hits the ground after approximately 5 seconds.