A bobcat jumps with a speed of 5 m/s at an angle of 55º. How long does the

bobcat stay in the air?

Answer - 0.84 s

y = y₀ + v₀ᵧ t + ½ gt²

where y is the final height,

y₀ is the initial height,

v₀ᵧ is the initial vertical velocity,

g is the acceleration due to gravity,

and t is time.

0 m = 0 m + (5 sin 55º m/s) t + ½ (-9.8 m/s²) t²

0 = 4.10 t - 4.9 t²

0 = t (4.10 - 4.9 t)

t = 0, t = 0.84

The bobcat lands after 0.84 seconds.