Missiles designed for defense against ballistic missiles hav

Missiles designed for defense against ballistic missiles have attained accelerations in excess of 100 g\'s, or 100 times the acceleration due to gravity. Suppose that the missile shown lifts off from the ground and has a constant acceleration of 100 g\'s. How long does it take to reach an altitude of 3000 m? How fast is it going when it reaches that altitude? Suppose that the missile shown lifts off from the ground and, because it becomes lighter as its fuel is expended, its acceleration (in g\'s) is given as a function of time in seconds by a = 100/1 - 0.2t. What is the missile\'s velocity in miles per hour 1 s after liftoff?

Solution

Using 2nd equation of Newton\'s law,
s=ut+1/2at^2
u=0
So,
s= 1/2 at^2
s = 3000
a=100
3000 = 1/2 * 100 *t^2
t^2 = 60
t = sqrt 60
t=7.75 sec

To find velocity when it reaches altitude,
v=u+at
v = 100 * 7.75
So, v= 775m/s.

4.) As a is variable,
so, v = adt
On integrating a,
v = -500(log t-5)
v = -693
So, v = 693m/s
v = 2494 km/hr