Question 1087514
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An arrow is shot vertically with an initial speed of 25m/sec can be approximated by the function h(t)=-5t^2+25t 
where h(t) is the height of the arrow (in meters) and t is the time after release of the arrow (in seconds) 


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a) how high is the arrow after 2.5 secs?                       - substitute t= 2.5 into h(t) and calculate.

b) how long does it take the arrow to return to the ground     - Find the positive root of the quadratic equation h(t) = 0.
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