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A stone is thrown at an angle of 30° above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A stop watch measures the stone's trajectory time from top of cliff to bottom to be 5.6 s. What is the height of the cliff? (g = 9.8 m/s² and air resistance is negligible)

Answer :

Answer:

the height was thrown = 120 m

Explanation:

given,

speed of the rock = 12 m/s

angle of launch = 30.0 ∘

horizontal distance = d = 15.5 m

acceleration due to gravity  = 9.8 m/s²

[tex]u_y = u sin \theta[/tex]

[tex]u_y = 12 sin 30^0[/tex]

[tex]u_y = 6 m/s[/tex]

t =5.6

[tex]s_y = u_y t + \dfrac{1}{2}gt^2[/tex]

[tex]s_y = 6\times 5.6 - \dfrac{1}{2}\times 9.8 \times 5.6^2[/tex]

[tex]s_y = -120.064m[/tex]

hence, the height was thrown = 120 m

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