physics
posted by Bob on .
A quarterback throws the football to a stationary receiver who is 18.3 m down the field. Football is thrown at an angle of 44.9 degrees to the ground. Acceleration is 9.81 m/s2. What initial spped must the quarterback throw ball to reach receiver and what is the highest point during it's flight.

The horizontal velocity is Vi*cos44.9 and the vertical velocity is Vi*sin44.9
horizontal
18.3=Vi*cos44.9 * time find time in terms of Vi.
Vertical
hf=hi+Vi*sin44.9*t 1/2 g t^2
now hf=hi=0
so
0=Vi*sin44.9 1/2 g t and put the expression you got for t, then solve for Vi. You will encounter the need for one trig identity. 
Split the problem in two, rising and falling. It does half the time and distance rising. (parabolic path)
Initial speed = S
U = constant speed in x direction = S cos 44.9 = .708 s
Vo = initial speed in y direction = S sin 44.9 = .706 s
vertical speed at max height = 0
so
0 = Vo  9.81 t
where t is time rising (half of total time in the air)
t = .706 s/9.81 = .0720 s
In half the time in the air (t) it goes half the distance down field , 9.15 m
U t = 9.15
t = 9.15 /(.708 s)
so
.072 s = 9.15 / (.708 s)
.051 s^2 = 9.15
s^2 = 180
s = 13.4 m/s
now how high?
t = 9.15/(.708 s) = 9.15/(.708*13.4) = .964 seconds rising
h = Vo t  4.9 t^2
h = .706*13.4 *.964  4.9 (.964^2)
= 4.56 meters (about 14 feet)