Question 274955
If an arrow is shot vertically upward with an initial velocity of 49 m/s, the height in metres reached in t seconds can be found the equation h=49t-4.9tsquared. Find the maximum height reached by the arrow.
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{{{h(t) = 49t - 4.9t^2}}}
When there's no 3rd term, the easiest way to do these is to find the time at apogee, which is 1/2 the total time.
Find the times when h= 0
{{{-4.9t^2 + 49t = 0}}}
t = 0 (at the start)
t = 10 (at impact)
The midpoint is the apogee, at t = 5 seconds
h(5) = 49*5 - 4.9*25
h max = 122.5 meters
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If the 3rd term is not zero(0), then find the vertex at t = -b/2a
t = -49/-9.8 = 5 seconds
same result