# When a truck is at x = 0, a cat jumps out of the back of the truck at an angle of 30 degrees with a velocity of 5 m/s relative to the truck. The back of the truck is 1.4 m above the ground. The...

When a truck is at x = 0, a cat jumps out of the back of the truck at an angle of 30 degrees with a velocity of 5 m/s relative to the truck. The back of the truck is 1.4 m above the ground. The truck is moving at 40 km/hr, where does the cat land?

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At x = 0, a cat jumps out of the back of the truck at an angle of 30 degrees with a velocity of 5 m/s relative to the truck. The back of the truck is 1.4 m above the ground. The truck is moving at 40 km/hr. The velocity of truck in m/s is 11.11 m/s. The cat jumps at a velocity 5 m/s relative to the truck.

The vertical component of the cat's velocity is 5*sin 30 = 2.5 m/s

The time t taken to travel a distance 1.4 m is given by -1.4 = 2.5*t - 4.9*t^2

This gives a positive solution approximately equal to 0.8474 s.

The horizontal velocity of the truck is 11.11 m/s. The cat jumps in the opposite direction giving a net horizontal velocity of the cat as 6.7798 m/s.

At this velocity the distance covered in 0.8474 s is 0.8474*6.7798 = 5.745 m

The cat lands at a spot that is 5.745 m from the point x = 0.