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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