Posted by Bob on Monday, January 19, 2009 at 6:47pm.
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.

physics  bobpursley, Monday, January 19, 2009 at 6:55pm
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.

physics  Damon, Monday, January 19, 2009 at 7:13pm
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)
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