Please help I am confused

Question

eyjafjallajökull is a volcano in ice land. during a recent eruption, the volcano, spewed out copious amounts of ash. one small was ejected from the volcano with an initial velocity of 336 ft/sec. the height h, in feet, of our ash projectile is given by the equation h= -16t2 +336t. where t is the time is seconds, assuming the volcano has no height. h=0 t=0 1. when does the ash projectile reach its maximum height? 2. what is its maximum height? 3. when does the ash projectile return to the ground

Answer 1

g=v/t

32.2=336/t

t=336/32.2=10.43sec

h=-16t^2+vt+s

=-16*10.43^2+336*10.43

=-1742.15+3506.08=1763.93ft

drop time= (t^2+t)/2*32.2=1763.93=9.98sec

Answer 2

For a quadratic function, y = Ax² + Bx + C, the x-coordinate of the vertex is -B / (2A). If A>0, the vertex is the lowest point on the graph and if A<0 the vertex is the highest point on the graph.

So, if H(t) = -16t² + 336t, then the maximum value occurs when t = -336 / (2(-16)) = 10.5.

Maximum height = H(10.5) = -16(10.5)² + 336(10.5) = 1764 ft

The ash projectile reaches the ground when H(t) = 0

So, -16t² + 336t = 0

-16t(t - 21) = 0

The ash projectile reaches the ground when t = 21 sec.

Answer 3

Since h= -16t² +336t then vel= -32t + 336. The particle its maximum when the velocity is 0. So 0=-32t +336 and t= 10.5 sec. The height h is -16(10.5)² +336(10.5) and h= 1774 ft The projectile returns to ground in twice the time to get to the maximum height and t =21 sec.