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A rock is thrown from a 50.0-m-high cliff with an initial velocity of 7.0 m/s at an angle of 53.0 degrees above the horizontal. How far from the base of the cliff will the rock hit?

Respuesta :

abiolaraymond1995

Answer:

4.8063m

Explanation:

Horzontal range is given by the formula;

R=(u²sin2θ)/g

u=7m/s,  θ=53°, g=9.8m/s

[tex]R=\frac{7^{2}sin2*53 }{9.8}[/tex]

[tex]R=\frac{7^{2}sin106 }{9.8}[/tex]

[tex]R=\frac{49*sin2*53 }{9.8}[/tex]

[tex]R=\frac{47.102 }{9.8}[/tex]

R=4.8063m

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