Solution :
The motion in the y direction.
The time taken by the toy rocket to hit the ground,
[tex]$S=ut+\frac{1}{2}at^2$[/tex]
S = distance travelled = 30 m
u = 0 m/s
a = [tex]$9.8 \ m/sec^2$[/tex]
t= time in seconds
Therefore, [tex]$30 =\frac{1}{2}9.8 t^2$[/tex]
t = 2.47 sec
Now motion in the x direction,
u = 12 m/sec
[tex]$a=1.6 t \ m/sec^3$[/tex]
Upon integration 'v' with respect to 't'
[tex]$v=\frac{1.6t^2}{2}+12$[/tex]
Once again integrating with respect to t,
[tex]$s=\frac{1.6t^3}{6}+12 t$[/tex]
[tex]$s=\frac{1.6(2.47)^3}{6}+12 (2.47)$[/tex]
= 0.0176+29.64
= 29.65 m
Therefore, the toy rocket will hit the ground at 29.65 m from the building.
