h=-4.9 t^{2}+25 tThe 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

Accepted Answer

Answer: D. 5.0 Choice \mathbf{D} is correct. When the ball hits the ground, its height is 0 meters. Substituting 0 for h in h=-4.9 t^{2}+25 t gives 0=-4.9 t^{2}+25 t, which can be rewritten as 0=t(-4.9 t+25). Thus, the possible values of t are t=0 and t=\frac{25}{4.9} \approx 5.1. The time t=0 seconds corresponds to the time the ball is launched from the ground, and the time t \approx 5.1 seconds corresponds to the time after launch that the ball hits the ground. Of the given choices, 5.0 seconds is closest to 5.1 seconds, so the ball returns to the ground approximately 5.0 seconds after

$$h=-4.9 t^{2}+25 t$$The 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?

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