Question

Some cats can be trained to jump from one location to another and perform other tricks. Kit the cat is going to jump through a hoop. He begins on a wicker cabinet at a height of 1.765 m above the floor and jumps through the center of a vertical hoop, reaching a peak height 3.130 m above the floor. (Assume the center of the hoop is at the peak height of the jump. Assume that +x axis is in the direction of the hoop from the cabinet and +y axis is up. Assume g = 9.81 m/s2.)

(a) With what initial velocity did Kit leave the cabinet if the hoop is at a horizontal distance of 1.560 m from the cabinet?
v_0 = m/s
(b) If Kit lands on a bed at a horizontal distance of 3.582 m from the cabinet, how high above the ground is the bed?
m

Answer 1

a. the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal

b. 0.847 m

Step-by-step explanation:

a. Using v² = u² + 2as, we find the initial vertical velocity of the cat. Now at the peak height, v = final velocity = 0, u = initial velocity and a = -g = 9.8 m/s², s vertical distance travelled by the cat from its position on the cabinet = Δy = 3.130 m - 1.765 m = 1.365 m.

Substituting these variables into the equation, we have

0² = u² + 2(-9.8m/s²) × 1.365 m

-u² = -26.754 m²/s²

u = √26.754 m²/s²

u = 5.17 m/s

To find its initial horizontal velocity, u₁ we first find the time t it takes to reach the peak height from

v = u + at. where the variables mean the same as above.

substituting the values, we have

0 = 5.17 m/s +(-9.8m/s²)t

-5.17 m/s = -9.8m/s²t

t = -5.17 m/s ÷ (-9.8m/s²)

= 0.53 s

Now, the horizontal distance d = u₁t = 1.560 m

u₁ = d/t = 1.560 m/0.53 s = 2.96 m/s

So, the initial velocity of the cat is V = √(u² + u₁²)

= √((5.17 m/s)² + (2.96 m/s)²)

= √(26.729(m/s)² + 8.762(m/s)²)

= √(35.491 (m/s)²)

= 5.95 m/s

its direction θ = tan⁻¹(5.17 m/s ÷ 2.96 m/s) = 60.2°

So, the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal

(b)

First, we find the time t' it takes the cat to land on the bed from d' = u₁t'

where d' = horizontal distance of cabinet from bed = 3.582 m

u₁ = horizontal velocity = 2.96 m/s

t' = d'/u₁

= 3.582 m/2.96 m/s

= 1.21 s

The vertical between the bed and cabinet which is the vertical distance moved by the cat is gotten from Δy = ut' +1/2at'²

substituting u = initial vertical velocity = 5.17 m/s, t' = 1.21 s and a = -g = -9.8 m/s² into Δy, we have

Δy = ut' +1/2at'² = 5.17 m/s × 1.21 s +1/2(- 9.8 m/s²) × (1.21 s)² = 6.256 - 7.174 = -0.918 m

Δy = y₂ - y₁

Since our initial position is the position of the cabinet above the ground = y₁ = 1.765 m

y₂ = position of bed above ground.

Δy = y₂ - y₁ = -0.918 m

y₂ - 1.765 m = -0.918 m

y₂ = 1.765 m - 0.918 m

= 0.847 m