| <<1 An athlete throws the shotput with an initial velocity of 14 m/s at a 40 degree angle from the horizontal. The shotput leaves his hand at a height of 2.2 m above the ground.
a. Determine the x and y components of the initial velocity.>>
The horizontal component is 14 cos 40 = 10.725 m/s and the vertical component is 14 sin 40 = 8.999 m/s.
<<b. Determine how long the shotput is in the air.>>
That is determined by the vertical component. The time t1 going up is such that g t = Vyo = 8.999 m/s. That means t1 = 0.92 s. In that time, the height increases to 2.2 m + (Vyo/2) t1 = 6.34 m. It spends a longer time t2 coming down because it was launched above the ground. (1/2) g (t2)^2 = 6.34 m. t2 = sqrt [2 * 6.34/g] = 1.14 s. Total time in air = t1 + t2 = 2.06s
<<c. Calculate the distance the shotput traveled.>>
Multiply the time in the air by the horizontal velocity component, which does not change. I get 22.09 m. | 
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