posted by andrew on .
1. Consider a projectile, say a frictionless teddy bear, which is thrown at an angle of theta with an initial
velocity of v0.
a) For a fi xed angle, and the maximum height of the teddy bear.
b) Calculate the distance that the teddy bear travels before returning (gently, of course) to the
ground (which is assumed to be
c) Use these pieces of information to nd the angle which produces the maximum of the sum of height and distance.
usage of function. in these question.
Vi = Vo sin T
u = Vo cos T
v = Vi - gt
max height when v = 0
0 = Vosin T - g t
t = Vo sin T / g
h = 0 + Vi t - .5 g t^2
max h = Vo sin T [Vo sin T/g] - .5 g[Vo sin T / g]^2
max h = [.5 /g][VosinT]^2
(this is quicker to get using potential and kinetic energy argumenents. Note max height of course when T = 90 degrees, straight up)
Now part b
the total time in the air is twice the time needed to reach max altitude so
total t = 2 Vo sin T/g
d = u t = Vo t cos T
d = Vo [ 2 Vo sinT/g]cos T
d = [2 Vo^2/g] sin T cos T
Sum = S = [.5 /g][VosinT]^2 + [2 Vo^2/g] sin T cos T
dS/dT = 0 at max
0= [.5Vo^2/g]sin Tcos T+[2Vo^2/g][-sin^2T+cos^2T]
0=2 cos^2T -.5 sin T cosT -2 sin^2T
but 2 (cos^2T-sin^2T)=2 cos2T
0= 2 cos 2T -.5 sinTcosT
but sinTcosT =.5 sin2T
0=2 cos2T -.25 sin2T
sin2T/cos2T = 8 = tan 2T
2T = 82.8 degrees
T = 41.4 degrees
Interesting, max range is at 45 degrees. I may have made an arithmetic error in that mess.
@Damon : Appreciate it sir. u are a life saver. i was on track and needed a kick . thanks alot.