5. Morgan leaves the long jump takeoff board with a vertical velocity of 2.8 m/s

Question

5. Morgan leaves the long jump takeoff board with a vertical velocity of 2.8 m/s and a horizontal 5m velocity of 7.7 m/s. () h2- a. What is Morgan's resultant velocity at takeoff? 2.6 7c18 59.24 b What is Morgan stake of angle the angle of her resultant takeoff velocity with- horizontal? What is Morgan's horizontal velocity just before she lands? If Morgan is in the air for 0.71 sec, what is her horizontal displacement during this time in the air? What is Morgan's vertical velocity at the end of her 0.71 sec flight? If Morgan's center of gravity was 1.0 m high at the instant of takeoff, how high will it be at the peak of her flight? How high is Morgan's center of gravity at the end of her flight, when she first hits the c. d. e. f. g. pit?

Explanation / Answer

a) since velocity is a vector quantity,

v = 7.7 i + 2.8 j

magnitude = (7.7^2+2.8^2)^0.5

=8.19 m/s

b)let the angle be x.so,

tan(x) = vy/vx

or tan(x) = 2.8/7.7

or tan(x)=0.3636

or x = 19.98 degrees with the horizontal

c)no force acts in the horizantal direction. hence the horizontal velcotiy remains the same.

so Vx = 7.7 m/s

d)Horizontal displacement = Vx*t

=7.7*0.71

=5.467 m

e)g = -9.81 m/s^2

so, Vf = Vi + 0.gt^2

=2.8-0.5*9.81*0.71^2

=0.32 m/s

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