h=−4.9t 2 +25tThe equation above expresses the approximate height ℎ, in meters, of a ball 𝑡𝑡 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?A3.5B4.0C4.5D5.0
Question
h=−4.9t 2 +25tThe equation above expresses the approximate height ℎ, in meters, of a ball 𝑡𝑡 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?A3.5B4.0C4.5D5.0
Solution
The ball hits the ground when the height (h) is zero. So, we need to solve the equation for t when h = 0.
0 = -4.9t^2 + 25t
This is a quadratic equation in the form of at^2 + bt + c = 0, where a = -4.9, b = 25, and c = 0.
We can solve for t using the quadratic formula: t = [-b ± sqrt(b^2 - 4ac)] / 2a
Substituting the values, we get:
t = [-25 ± sqrt((25)^2 - 4*(-4.9)0)] / 2(-4.9) t = [-25 ± sqrt(625)] / -9.8 t = [-25 ± 25] / -9.8
This gives us two solutions: t = 0 and t = 5.1
Since time cannot be negative, we discard t = 0 (which represents the initial launch of the ball).
Therefore, the ball will hit the ground after approximately 5.1 seconds. So, the closest answer is D5.0.
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