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school question? A long jumper travels 8.95 meters during a jump. He moves at 10.8 m/s when starts his leap. At what angle from...Question
A long jumper travels 8.95 meters during a jump. He moves at 10.8 m/s when starts his leap. At what angle from the horizontal must he have been moving when he started his jump? You may need the double-angle formula: 2 sin u cos u = sin (2u)
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sue
please can someone help?
drwls
The formula they are refering to is
X = (V^2/g) sin 2A
8.95 = 10.8^2/9.8 * sin@A = 11.9 sin2A
sin2A = 0.752
2A = 48.8 degrees or 131.2
A = 24.4 or 65.6 deg
The smaller angle is more likely because it allows the long jumper to get added distance by extending feet forward at landing
X = (V^2/g) sin 2A
8.95 = 10.8^2/9.8 * sin@A = 11.9 sin2A
sin2A = 0.752
2A = 48.8 degrees or 131.2
A = 24.4 or 65.6 deg
The smaller angle is more likely because it allows the long jumper to get added distance by extending feet forward at landing
sue
that is indded correct! can you help me on the other question i had too? its right above this post. it is either the electron or cannon question. thanks so much again!