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A student fires a cannonball vertically upwards. The cannonball returns to the ground after a 5.5s flight. Determine all unknowns and answer the following questions. Neglect drag and the initial height and horizontal motion of the cannonball. Use regular metric units (ie. meters).

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Answer:

The highest point of the cannonball displacement is h=37.06m, the time it uses to get there is t=2.75s, and its initial speed is
v_i=26.95m/s.

Explanation:

The time the cannonball uses to return is the same it uses to get to the highest point of its displacement, so the time for each trip is 5.5s/2=2.75s

Initial speed
v_i can be determined by:


v_f -v_i =-g(t_f -t_i )


0-v_i=-9.8(2.75-0)


v_i=26.95m/s

The height the cannonball reached is:


h=h_i+v_i t-(1)/(2)gt^2=0+26.95*2.75-(1)/(2)9.8(2.75)^2


h=37.06m

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