Some cats can be trained to jump from one location to another and perform other

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 1.786 m above the floor and jumps through the center of a vertical hoop, reaching a peak height 3.120 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/s^2.) With what initial velocity did Kit leave the cabinet if the hoop is at a horizontal distance of 1.552 m from the cabinet? (Express your answer in vector form. Give your answer to at least two decimal places for each vector component.) v^rightarrow_0 =___________m/s If Kit lands on a bed at a horizontal distance of 3.590 m from the cabinet, how high above the ground is the bed?__________m

Explanation / Answer

A.

height jumped by cat above the launching point = 3.120 - 1.786 = 1.334 m

at max height using kinetic equation

h = u*t + 0.5a*t^2

h = 0 + 0.5*g*t^2

t = sqrt(2*h/g)

t = sqrt(2*1.334/9.81) = 0.5215 sec

so time for the cat to reach at max height = 0.5215 m

So

initial vertical component of cat's velocity will be

Voy = g*t = 9.81*0.5215 = 5.115 m/sec

horizontal component of cat's velocity will be

Vox = 1.552/t = 1.552/0.5215 = 2.976 m/sec

So velocity will be

Vo = (2.976 i + 5.115 j ) m/sec

B.

time for cat to reach the bed = distance/velcoity

T = d/Vox

T = 3.590/2.976 = 1.206 sec

cat was free falling for the time t1 = T - t

t1 = 1.206 - 0.5215 = 0.6845 sec

vertical distance traveled after max height

H = 0.5*g*t1^2

H = 0.5*9.81*0.6845^2 = 2.298 m

So height of bed will be

H1 = h - H

H1 = 3.120 - 2.298 = 0.822 m

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