Projectile motion!! help!! eaxy math!?

Question

#1) The height of the projectile fired upward is given by the formula s=Vot-16t ^2, where s is the height and Vo is the initial velocity and t is time. Find time for a projectile to return to Earth if it has velocity of 200ft/s.

I know in this one, i solve for t, but distance would I use for s?

#2) . In Germany, there are no speed limits on some portions of highway. Others have a speed limit of 180 km/h (approximately 112 mph.) required to stop a car traveling at v kilometers per hour is d=0.019v^2+0.69v. Approximate to nearest tenth.

I don't get this one at all!!

#3) A model rocket is launched with an intial velocity of 200ft/s. The height h, of the rocket t seconds after the launch is given by h= -16ft^+200t. How many seconds after launch will rocket be 300 ft above ground? Round nearest hundredth of a sec.

Answer 1

1) the projectile has an initial velocity upwards of 200 f/s, gravity is accelerating the object downwards at 16f/s^2, and the object starts at ground level (s=0) so 0=200t-16t^2. solving for t gives you the time it takes the projectile to reach its maximum height at which point it heads downwards. the time it takes to reach earth again will be the same time it took to get to the maximum height so the total time (T) will be equal to two times the time to get to max height(t). T=2t

2) im not really sure what you're asking to figure out. stopping distance? it looks to me like all the information is there so that all you need to do is plug in the velocity (180km/h) into the function: d=(.0019)(180)^2+(.69)(180)

3) h=300 so 300=-16t^2+200t
adjust the equation so that it reads 16t^2-200t+300=0
either figure out the roots by simplifying to 4t^2-50t+75 or by plugging into a calculator and finding where the function crosses the x-axis
there will be two times at which the height will be 300 ( on the way up and on the way down) unless the height is a maximum

Answer 2

condition for OP to be perpendicular to the path of action at P If equation of inverted parabola is of type y = mx - nx^2 dy/dx = m - 2nx Slope at x = 0 is only m = tan a Slope OP could settle for by employing y/x = m - nx condition for perpendicularity is (m - 2nx)(m - nx) = -a million (2n^2)x^2 - (3mn)x + (m^2 +a million) = 0 x = [3m +/- sqrt(m^2 - 8)]/4n You frequently get 2 consequences. working example y = 3x - x^2 x = 2 or 5/2 This eqn shows m^2 >/= 8 m = tan a is >/= 2sqrt(2) ~ 2.80 3 Regards - Ian