Aprojectile is launched upward from ground level with an intentional speed of 98 m/s. how high will it go? when will it return to the ground?

Let h(t) represent the height of the projectile after t seconds. If the projectile is being launched from the ground, then h(0)=0, and h(t)=-4.9t²+98t.  When the equation is in this form, the maximum height is determined by -98/2(-4.9), or after 10 seconds. Thus
-4.9(100)+98(10)=490 m as the highest point of the projectile
Then, when it hits the ground, h(t)=0, so
0=-4.9t²+98t
98t=4.9t²
4.9t=98
t=20
☺☺☺☺