KJF2nd 3-65 In a shot put event, an athlete throws the shot with an initial velo
ID: 1879745 • Letter: K
Question
KJF2nd 3-65 In a shot put event, an athlete throws the shot with an initial velocity of i 12.0.m/s at a 40.0° angle above the horizontal. The shot leaves her hand at a height of 1.80 m above the ground. Ignore air resistance. a) How long is the shot in the air? (1.78s) b) How far horizontally does the shot travel? (16.4 m) c) What is the maximum height of the shot above the ground? (4.84 m) d) How long does it take for the shot to reach its maximum height? (0.787 s) e) CAREFULLY sketch the horizontal and vertical components of the shot's velocity as a function of time. Label all axes with numerical values. f CAREFULLY sketch the horizontal and vertical position of the shot as a function of time. Label all axes with numerical values. m /' '--+ 1 h. aa 2 y--Explanation / Answer
a] Time taken for the put to reach the maximum height:
t = uy/g = (12sin40)/9.8 = 0.787s
so, the total time for which the put remains in the air is: T = 2t + t' = 1.574 + t'
where t' is the time it takes from the point it reaches again the height of 1.8m to when it reaches the ground.
1.8 = 12sin40 t' + (1/2)(9.8)t'2
=> 4.9t'2 + 7.713 t' - 1.8 = 0
solving this quadratic equation gives, t' = 0.206s
therefore, T = 1.574 + 0.206 = 1.780 s.
b] L = uxT = 12cos40 x 1.78 = 16.37 m.
c] at the highest point, vy = 0 m/s
vy = uy - gt
t = uy/g = 0.787s.
d] vy = 12sin40 - 9.8t
vx = 12cos40
e] y = 12sin40 t - (1/2)9.8t2
x = 12cos 40 t .
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