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4 years ago

12

A worker kicks a flat rock lying on a roof. The rock slides up the incline 10.0 m to the apex of the roof, and flies off the roo

f as a projectile. What maximum height (in m) does the rock attain? Assume air resistance is negligible, vi = 15.0 m/s, μk = 0.425, and that the roof makes an angle of θ = 42.0° with the horizontal. (Assume the worker is standing at y = 0 when the rock is kicked.)

1 answer:

masya89 [10]4 years ago

7 0

Answer:

h = 7.42 m

Explanation:

deceleration of the rock

$a = g sin \theta + \mu_kgcos \theta$

$a = 9.8 sin 42^0+0.425\times 9.8 \times cos 42^0$

a = 9.65 m/s²

using formula

v² = u² + 2 a s

v² = 15² - 2×9.65 × 10

v = 5.66 m/s

the height attained is

h₁ = 10 sin θ

= 10 sin 42

= 6.69 m

now with vertical velocity it will reach to the height h₂

v y = v sin θ

= 5.66 sin 42

= 3.79 m/s

height is

v² = u² + 2 a s

0 = 3.79² - 2 × 9.8 ×h₂

h₂ = 0.73 m

the maximum height is

h = h₁ + h₂

= 6.69 + 0.73

h = 7.42 m

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