a) when h(t)=0 the arrow is on the ground, so 5t²-29t-2=0. Using the quadratic formula we get t=(2.9±√881/10).
This gives us two values for t. The difference between the two values is the length of time the arrow is in the air=√881/5=5.94 seconds.
b) the maximum height is when the speed of the arrow is zero, which is given by h’(t)=-10t+29=0, so t=2.9 seconds.
h(2.9)=44.05m.
c) -5t²+29t+2≥3, 5t²-29t+1≤0. Using the quadratic formula again, solve the equality before the inequality:
t=2.9±√821/10 seconds. The graph is an upright U-shaped parabola cutting the t axis at 0.0347 and 5.7653. Therefore between these times the graph is below the t axis where h(t)≥3, meaning that the arrow is at least 3m off the ground.
