Question

A long jumper leaves the takeoff board with her center of mass vertical velocity of 3.7 m/s and horizontal velocity of 6.5 m/s.

a). What is the long jumper’s resultant velocity at takeoff?

b) What is the takeoff angle?

c) What is the horizontal velocity just prior to landing?

d) If the jumper is in the air for 0.75 s, what is her horizontal displacement during the time in the air?

e) What is the jumper’s vertical velocity at the end of the 0.75 s flight?

Answer 1

a) Vertical Velocity = 3.7m/s

Horizontal velocity = 6.5m/s

Resultant velocity magnitude = = 7.47 m/s

b) Takeoff angle = tan^-1(3.7/6.5) = 29.63 degrees from horizontal

c) Just prior to landing: Horizontal Veocity = 6.5m/s ( since no acceleration in horizontal direction)

d) Horizontal displacement = Horizontal velocity*time = 6.5*0.75 = 4.875m

e) Vertical velocity(@0.75s) = 3.7 - 9.81*0.75 =  -3.6575 m/s (V = u + at) where a = -g (acceleration due to gravity = 9.81m/s²)