Answer:
The rocket would be in the air for 8.125 seconds after launch.
Step-by-step explanation:
Given that the height of the rocket, h = 260t - 16[tex]t^{2}[/tex].
To determine how long the rocket would be in the air, differentiate h totally with respect to t.
So that, total derivative of h is given as;
[tex]\frac{dh}{dt}[/tex] = [tex]\frac{d(260t-16t^{2}) }{dt}[/tex]
0 = 260 - 16 x 2t
= 260 - 32t
0 = 260 - 32t
32t = 260
t = [tex]\frac{260}{32}[/tex]
= 8.125
t= 8.125 seconds
The rocket would be in the air for 8.125 seconds after launch.
