A projectile is fired straight up from ground level. After t seconds its height s, in feet above the ground, is given by s=350t-16t^2.

Question

A projectile is fired straight up from ground level. After t seconds its height s, in feet above the ground, is given by s=350t-16t^2. During what time interval is the projectile more than 1250 ft above the ground?

Answer 1

Since s=350t-16t², we want to know the 2 times where 350t-16t²=1250, as those two values will be the begin and end times of the interval in question. We can rewrite that equation in standard quadratic equation form:

16t²-350t+1250=0 or, simplified, 8t²-175t+625=0

Since it is now in quadratic form, we can solve using the quadratic formula:

t = (175±√((-175)²-(4*8*625)))/(2*8)=(175±√(30625-20000))/16=(175±√10625)/16≈(175±103)/16

So the time period from approximately 72/16 to 178/16 or 4.5-11.125 seconds has the bullet above 1250 feet.

Answer 2

We can just attack this problem with solving the given equation on substituting 1250 feet for s to obtain t which will most likely lead to solving for the roots of the quadratic in t which is somewhat ugly.

We can also use what we should know by now concerning the motion of masses experiencing uniform acceleration. Here the acceleration is that of gravity 32 feet per second per second.

The expression we have been given can then be interpreted for an initial velocity of the projectile to be 350 feet per second. What is the initial energy of the projectile? That is the kinetic energy of motion which is 1/2 mass times (350)² foot-lbs. On traveling to 1250 feet, some of this energy has been converted to potential energy mass x 32 feet per second per second x 1250 feet. The change in energy is equivalent to the new kinetic energy of the projectile on reaching 1250 feet that is associated with a new velocity given by the relation v= 350 ft/sec - 32 ft/sec-sec x elapsed time (sec). This relation enables analysis of this problem avoiding the quadratic.

Here the elapsed time is given on rearrangement as [350 ft/sec - v(t)]/32.

Let's simplify this a bit further, consider the apogee of the projectile motion, now v(t) is zero and the time elapsed to reach this point is(350 ft/sec)/32 ft/sec-sec) = 10.9375 seconds.

Now, we can find the height at apogee using our original equation s=350 ft/sec x 10.9375 sec - 16 ft/sec-sec x (10.9375 sec)²

This height is 3828.125 ft - 1914.0625 ft = 1914.0625 ft

Back to looking at changes in energy...the potential energy of our projectile at apogee is mass x 32 ft/sec-sec x 1914.0625 ft. The potential energy of our projectile at 1250 ft above the ground is mass x 32 ft/sec-sec x 1250 ft the difference in these energies is the kinetic energy of the projectile both on its way up and way down which is 1/2 mass x v(t)² so we now have 2 x 32 ft/sec-sec ( 1914.0625 ft - 1250 ft) = v(t)² and we don't need to know the mass of our projectile to perform this analysis or v(t) is the SQRT( 2 times g times the difference in the projectile heights).

More generally we have seen in kinematics the equation SQRT(2as) = v(t) - v_o developed to eliminate time in relating distance traveled to final velocity and initial velocity.

So now what is the projectile velocity on reaching 1250 ft...that is SQRT(42500) = 206.16 ft/sec.

And how much time must elapse for the velocity of the projectile at apogee to now reach a velocity of 206.16 ft/sec...that is just 32ft/sec-sec divided into 205.16 ft/sec or 6.44 sec.

Now we have the full story; the projectile travels to apogee in 10.9375 seconds and remains above 1250 ft for 6.44 seconds. The total time of flight is twice the time to apogee 22 seconds. the projectile is more than 1250 ft above its launch point for 12.9 seconds- 6.44 seconds on the way up and 6.44 seconds on the way down.