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

t= 2vₒ•sinα/g

6 seconds

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

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To determine how long the bobcat stays in the air, we need to find the time it takes for the bobcat to reach the ground.

First, let's break down the problem into its horizontal and vertical components. The initial speed of the bobcat can be split into two components: the horizontal component (Vy) and the vertical component (Vx).
Given that the bobcat jumps at an angle of 55 degrees, we can use trigonometry to find the horizontal and vertical components of its velocity.

The horizontal component (Vx) can be found using the equation:
Vx = V * cos(angle)

The vertical component (Vy) can be found using:
Vy = V * sin(angle)

Substituting the given values into the equations:
Vx = 5 m/s * cos(55 degrees)
Vx ≈ 5 m/s * 0.5736
Vx ≈ 2.868 m/s

Vy = 5 m/s * sin(55 degrees)
Vy ≈ 5 m/s * 0.8192
Vy ≈ 4.096 m/s

Now, we can analyze the vertical motion of the bobcat and determine how long it takes to reach the ground. We can use the equation for vertical displacement:
h = (Vy * t) - (0.5 * g * t^2)

Where:
h = vertical displacement (0 in this case, as the bobcat starts and ends on the ground)
Vy = vertical component of velocity (4.096 m/s)
t = time in seconds
g = acceleration due to gravity (approximated as 9.8 m/s^2)

Since the bobcat starts and ends on the ground, the vertical displacement (h) is zero. Solving the equation for t:

0 = (4.096 m/s * t) - (0.5 * 9.8 m/s^2 * t^2)

Rearranging the equation:

4.9 t^2 - 4.096 t = 0

Factoring out t:

t (4.9t - 4.096) = 0

Solving the equation, we have two possible solutions:
1) t = 0 (not feasible in this case)
2) t = 4.096 m/s / 4.9 m/s^2

Calculating the time t:

t = 4.096 m/s / 4.9 m/s^2
t ≈ 0.836 seconds

Therefore, the bobcat stays in the air for approximately 0.836 seconds.