Calculus - Distance of a Cannon Ball

Question

Suppose we fire a cannonball from the point (0,0) on the xy-plane at the angle of 60 degrees from the x-axis. Assume that the initial speed of the cannonball is 80 m/s, and the acceleration due to gravity is −10 m/s^2. How far does the cannonball travel?

Answer 1

at 60 degrees, and vo=80 m/s

initial vertical velocity= (sqr3/2)80= 40sqr3

height = h(t)=-(10/2)t^2 +(40sqr3)(t)

max h is when h'=0 = -10t +40sqr3

t=4sqr3

double that time

it hits the ground in 8sqr3 seconds= about 8(1.732) seconds

its trajectory is a parabola

with x intercepts 0 and 8sqr3

axis of symmetry t=4sqr3

h(4sqr3)= -5(4sqr3)^2+40sqr3(4sqr3)

=240+480= 240 meters = maximum height

horizontal initial velocity= 80/2 = 40 m/s

40 x 8sqr3 = 320sqr meters

horizontal acceleration = -5 m/s^2

s(t) = -2.5t^2 -40t

s(8sqr3) =- 2.5(8sqr3)^2 + 40(8sqr3)

= -7.5(64) +320sqr3

= -480 + 320sqr3

= about -320(1.732)-480 meters

=about 554.24-480

=about 74.24 meters, if accerlation horizontally = -2,5t^2

or about 1,034.24 meters, if the acceration =+2.5t^2 horizontally

which is it?