physics
posted by kristy .
Suppose that a shot putter can put a shot at the worldclass speed v0 = 14.00 m/s and at a height of 2.160 m. What horizontal distance would the shot travel if the launch angle θ0 is (a) 45.00° and (b) 42.00°? The answers indicate that the angle of 45°, which maximizes the range of projectile motion, does not maximize the horizontal distance when the launch and landing are at different heights.

a. Vo = 14m/s[45o]
Xo = 14*cos45 = 9.9 m/s.
Yo = 14*sin45 = 9.9 m/s.
Y = Yo + g*t = 0
Tr=Yo/g = 9.9/9.8 = 1.01 s. = Rise time.
hmax = ho + Yo*t + 0.5g*t^2
hmax=2.16 + 9.9*1.01  4.9*1.01^2=7.16m
Above gnd.
h = 0.5g*t^2 = 7.16
4.9t^2 = 7.16
t^2 = 1.46
Tf = 1.21 s. = Fall time.
D = Xo * (Tr+Tf)=9.9 * (1.01+1.21=21.9m
= Hor. distance.
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