bob beamon, an Olympic long jumper, can jump 8.90 meters. How long will he be in the air and how high will he travel if his horizontal speed is 9.1 m/s when he leaves the ground? Assume he lanfs upright, just as he did when he first hit the ground.
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Given : can jump 8.90 meters
horizontal speed is 9.1 m/s
To Find : How long will he be in the air
how high will he travel
Solution:
Horizontal Distance = 8.9 m
Horizontal Speed = 9.1 m/s
Time of Flight = 8.9/9.1 = 0.978 sec
He will be in air for 0.978 secs
Time for max height = 0.978/2 = 0.489 secs
V = U + at
Vertical velocity at top is 0
a = -g = -9.8 m/s²
0 = U -9.8*0.489
=> U = 4.7922 m/s
Initial Vertical velocity = 4.7922 m/s
V² - U² = 2aS
=> 0² - (4.7922)² = 2(-9.8)S
=> S = 1.1716929 m
He will be about 1.1717 m high
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