A cannonball was fired horizontally 47.3 m at a speed of 2.6 m/s. How high was the cannon?

Dxғ = 47.3 m = final horizontal displacement

Vxi = 2.6 m/sec = initial horizontal velocity

h{t} = 0 = height at time "t" (impact with the ground)

Hi = initial height

Vyi = 0 = initial vertical velocity


   Dxғ = (Vxi) • t   ...   horizontal (constant velocity) equation
   47.3 = (2.6) • t
        t = 18.19 sec   ...   time to impact with the ground

       h{t}  =  (½) • g • t²   +   (Vyi) • t   +   Hi   ...   vertical acceleration due to gravity equation
         0  =  (½) • (-9.8) •(18.19)²   +   0 • (18.19)   +   Hi
         Hi  =  1621.7 m   ...   initial height

t=18.2 s
h = 4.9(18.2)^2=1623 m

d/v = (2y/g)^1/2

d^2 / v^2 = 2y/g

y = 1/2 * d^2 g / v^2
= 1/2 * (47.3 m)^2 * (9.81 m/s2) / (2.6 m/s)^2

= 1600 m

About 5 beers.

Really, really high. It just got done hanging out with Snoop Dogg