A small dog is trained to jump straight up a distance of 1.2 m. How much kinetic energy does the 7.2-kg dog need to jump this high? (The acceleration due to gravity is 9.8 m/s².)

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Sanchez Mr Potato? Q

To find the kinetic energy required for the small dog to jump this high, we need to know the initial velocity of the dog just before it jumps.

The initial velocity can be determined using the following kinematic equation:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s as the dog reaches its highest point)
u = initial velocity (unknown)
a = acceleration due to gravity (-9.8 m/s² because it's opposite to the direction of motion)
s = displacement (1.2 m)

Rearranging the equation, we have:

u^2 = v^2 - 2as

Plugging in the known values:

u^2 = 0 - 2(-9.8)(1.2)
u^2 = 23.04
u ≈ √23.04
u ≈ 4.80 m/s

Now, with the initial velocity, we can calculate the kinetic energy using the formula:

KE = 0.5mv^2

Where:
KE = kinetic energy
m = mass of the dog (7.2 kg)
v = velocity (4.80 m/s)

Plugging in the known values:

KE = 0.5 * 7.2 * (4.80)^2
KE ≈ 0.5 * 7.2 * 23.04
KE ≈ 82.944 J

Therefore, the small dog needs approximately 82.944 Joules of kinetic energy to jump a distance of 1.2 meters.