Vertical Motion

Question

Katy twirls baton for Howard Huskies Marching band, and she releases her baton into the air when it is 6 feet above the ground. The initial velocity of the baton is 37 feet per second. Katy will catch the baton when it falls back to the height of 7 feet. How long is the baton in the air (in seconds)?

Answer 1

height=Vo^2/(2g)

=37fs^2/(2*32.2fs)

=1369/64.4

=21.257ft

gravity distance down=(t^2+t)/2*32.2

This is to drop until it hits 7'; so (t^2+t)/2*32.2=(14.257).

t=.566 sec

Answer 2

The formula is for projectile motion is

h=-16t²+v_ot+h_o

v_o is the initial velocity and h_o is the initial height. Plugging in the values given:

7=-16t²+37t+6 Subtract 7 from both sides

0=-16t²+37t-1

Solving using the quadratic equation, 2.285 seconds the baton is in the air