Question

A toy rocket is launched with an initial velocity of 12.0 m/s in the horizontal direction from the roof of a 30.0-m-tall building. The rocket’s engine produces a horizontal acceleration of \left(1.60 m / s^{3}\right) t, in the same direction as the initial velocity, but in the vertical direction the acceleration is g , down-ward. Ignore air resistance. What horizontal distance does the rocket travel before reaching the ground?

Answer 1

Solution :

The motion in the y direction.

The time taken by the toy rocket to hit the ground,

$S=ut+\frac{1}{2}at^2$

S = distance travelled = 30 m

u = 0 m/s

a = $9.8 \ m/sec^2$

t= time in seconds

Therefore, $30 =\frac{1}{2}9.8 t^2$

t = 2.47 sec

Now motion in the x direction,

u = 12 m/sec

$a=1.6 t \ m/sec^3$

Upon integration 'v' with respect to 't'

$v=\frac{1.6t^2}{2}+12$

Once again integrating with respect to t,

$s=\frac{1.6t^3}{6}+12 t$

$s=\frac{1.6(2.47)^3}{6}+12 (2.47)$

  = 0.0176+29.64

   = 29.65 m

Therefore, the toy rocket will hit the ground at 29.65 m from the building.