V = Vi up and u horizontal
u = V cos 30 = .866 V
Vi = V sin 30 = .5 V
in air t seconds
horizontal problem:
12.2 = u t
12.2 = .866 V t
V t = 14.1
so 2Vi t = 14.1 and Vi t = 7.05
vertical problem
h = Vi t - 4.9 t^2 = 0 at start and finish
0 = 7.05 - 4.9 t^2
t = 1.2 seconds in air
V =14.1/t = 14.1/1.2 = 11.75
agree with you
max height at t/2 or at 0.6 seconds
h = 3.5 - 4.9(.36)
about 1.75 meters
The quarterback of a football team gets off a perfect pass at an angle of 30 degrees with the horizontal. His receiver gathers it in 12.2m downfield from the passer at the same height above ground as the passer's throwing hand.
a. What was the ball's velocity as it left the passer?
b. How high above the passer's outstretched arm did it rise?
For part a. I got a answer of 11.7m/s by using the substitution method. Is this right?
I'm having a bit of trouble finding part b. I'm using the equation s=Vi*t+1/2a*t^2
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