# A bobcat jumps with a speed of 5 m/s at an angle of 55°. How long does the bobcat stay in the air?

Sep 26, 2017

$1.02 s$

#### Explanation:

The angle and mass are irrelevant to the "hang time", although they would be required for the distance or trajectory.

We only need to calculate the time for the initial velocity to match the gravitational pull (acceleration). The round trip time is twice the time to the peak.
$g = 9.8 \frac{m}{s} ^ 2$
$\frac{5 \frac{m}{s}}{9.8 \frac{m}{s} ^ 2} = 0.51 s$ Going UP!
$0.51 \times 2 = 1.02 s$ Round Trip.

Sep 26, 2017

0.835 s

#### Explanation:

By eliminating the time of flight from the vertical and horizontal components of motion we get this expression for the range d of the projectile:

$\textsf{d = \frac{{v}^{2} \sin 2 \theta}{g}}$

The horizontal component of the velocity is a constant $\textsf{v \cos \theta}$.

So we can say $\textsf{d = v \cos \theta t}$.

Putting the two expressions for d equal to each other we get:

$\textsf{\frac{{v}^{\cancel{2}} \sin 2 \theta}{g} = \cancel{v} \cos \theta t}$

$\therefore$$\textsf{t = \frac{v \sin 2 \theta}{g \cos \theta}}$

$\textsf{t = \frac{v 2 \sin \theta \cancel{\cos \theta}}{g \cancel{\cos \theta}}}$

$\textsf{t = \frac{5 \times 2 \sin 55}{9.81} = 0.835 \textcolor{w h i t e}{x} s}$