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The path of a stunt car driven horizontally off a cliff is represented by the diagram below. After leaving the cliff, the car falls freely to a point A in 0.50 second and to the point B in 1.00 second​

The path of a stunt car driven horizontally off a cliff is represented by the diagram below After leaving the cliff the car falls freely to a point A in 050 sec class=

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onyebuchinnaji

The horizontal component of the velocity at point B is 2.45 m/s.

Height of the car at point A

h(A) = v + ¹/₂gt^2

h(A) = v + (0.5 x 9.8)t²

h(A) = v + 4.9t²

h(A) = v + 4.9 x (0.5)²

h(A) = v + 1.225

Height of the car at point B

h(B) = v + ¹/₂gt^2

h(B) = v + 0.5 x 9.8 x (1)²

h(B) = v + 4.9

Horizontal component of the velocity is constant,

h(A)/t₁ = h(B)/t₂

[tex]\frac{v + 1.225}{0.5} = \frac{v + 4.9}{1} \\\\v + 1.225 = 0.5(v + 4.9)\\\\v + 1.225 = 0.5v + 2.45\\\\0.5v = 1.225\\\\v = \frac{1.225}{0.5} \\\\v = 2.45 \ m/s[/tex]

Thus, the horizontal component of the velocity at point B is 2.45 m/s.

Learn more about horizontal component of velocity here: https://brainly.com/question/24949996

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