Acannon fires a projectile as shown .the dashed line shows the trajectory in the absence of gravity; points mnop correspond to the position of the projectile at one second intervals. if g =10m/s 2 ,the lengths x, y,z are:

X = 5m, Y = 20m, Z = 45m

Explanation:

1) See attached the figure of this problem (missing in your question).

2) The equation that models the height with initial velocity Voy, in absence of gravity (dashed line) is:

H = Voy × t

3) The equation that models the height with initial velocity Voy and with gravity (parabola) is:

H' = Voy×t - g×t² / 2

4) The lengths requested correspond to the diffference in heights:

H - H' = g×t² / 2

5) Using g = 10 m/s²

H- H' = 5t²

6) X, Y and Z are at t = 1 s, t = 2 s, and t = 3 s

X = 5m/s² × (1s)² = 5 m

Y = 5m/s² × (2s)² = 20 m

Z = 5m/s² × (3s)² = 45 m