Suppose that a shot putter can put a shot at the worldclass speed 15.00 m/s and at a height of 2.160 m. What horizontal distance would the shot travel if the launch angle 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.
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Answer:
Given:
v
0
=15.00m/s
y
0
=2.160m
Then,
(a) y−y
0
=(v
0
sinθ
0
)t−
2
1
gt
2
⇒0−2.16m=(15.00sin45
∘
m/s)t−
2
1
(9.80m/s
2
)t
2
⇒4.9 t
2
−10.61 t−2.16=0
⇒t=−0.19s or t=2.35s
Thus, the total travel time of the shot in the air is found to be t=2.35s.
Therefore, the horizontal distance traveled is
R=(v
0
sinθ
0
)t=(15.00m/s)cos45
∘
×2.35s=24.95m
(b) Using the procedure outlined in (a) but for θ
0
=42
∘
, we have:
y−y
0
=(v
0
sinθ
0
)t−
2
1
gt
2
⇒0−2.16m=(15.00sin42
∘
m/s)t−
2
1
(9.80m/s
2
)t
2
⇒4.9 t
2
−10.04 t−2.16=0
⇒t=−0.20s or t=2.24s
Thus, the total travel time of the shot in the air is found to be t=2.35s.
This gives:
R=(v
0
sinθ
0
)t=(15.00m/s)cos42
∘
×2.24s=25.02m
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