an aeroplane moving horizontally with a speed of 180 km/hr drops a food packet while flying at a height of 490m . the horizontal range of the packet is??

Question

anonymous · Answer

$$t = \sqrt{2\times490/9.81}$$
t = 10 seconds

speed = 180/3.6 = 50m/s

10*50 = 500m

anonymous · Answer

This is effectively a projectile motion with teh initial conditions that the projectile is launched at a n angle of zero to teh horizontal.  We can separate the initial velocity of the object into its horizontal and vertical components.  We know that its horizontal component will be 180 km/hour (which is 50 m/s).  We know that the vertical component is zero.

So the first thing we need to know is the time $t$ it takes for the package to hit the ground.  This can be determined from the equation of motion $$s=u_vt+\frac{1}{2}at^2$$where $s$ is the initial height, $u_v$ the initial velocity, and $a$ is the acceleration due to gravity.  If $s=490$m and $u_v=0$ and assuming $a=9.8$ we have (after rearranging) $$t=\sqrt{\frac{(2)(490)}{9.8}}=10$$

We can now calculate the range of the object by considering the horizontal component of the initial velocity, remembering that because it is horizontal, it will be constant (as gravity only effects the vertical component).  The range then will be the distance the object travels horizontally from the time of release to the time of hitting the ground.  We have just worked out the time it takes to hit the ground to be 10 seconds.  So we know the initial speed horizontally $u_h=50$, and we know that the range of an object is the product of its distance and time of flight, or $$R=u_ht$$.   If you wonder where this comes from recall the equation of motion above (but considering the horizontal component of initial velocity $u_h$ $$s=u_ht+\frac{1}{2}at^2$$and remembering that $a=0$ in the horizontal direction, and thus we are left with $$R=s=u_ht$$Plug the numbers in and we get that the range is $$R=50\times10=500\rm{m}$$which is the same as the above answer.