Answer:
t = 15 s
Step-by-step explanation:
A projectile is thrown upward so that it's distance above the ground after t seconds is as follows :
[tex]h=-16t^2+480t[/tex] ...(1)
We need to find after how many seconds does it reach its maximum height. For maximum height, put [tex]\dfrac{dh}{dt}=0[/tex]
[tex]\dfrac{d(-16t^2+480t)}{dt}=0\\\\-32t+480=0\\\\t=\dfrac{480}{32}\\\\t=15\ s[/tex]
So, it will reach its maximum height in 15 seconds.
