Answer:
h = 7.42 m
Explanation:
deceleration of the rock
[tex]a = g sin \theta + \mu_kgcos \theta[/tex]
[tex]a = 9.8 sin 42^0+0.425\times 9.8 \times cos 42^0[/tex]
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
The maximum height that the rock attains is : 7.42 m
Given data :
Vi = 15.0 m/s
uk = 0.425
angle made by roof ( θ ) = 42.0°
First step : calculate the deceleration of the rock
a = g*sinθ + uk * g * cosθ
= 9.8 * sin 42° + 0.425 * 9.8 * cos 42°
= 9.65 m/s²
Applying motion formula
v² = u² + 2as
= 15² - 2 * 9.65 * 10
v = 5.66 m/s
Initial height attained ( h₁ ) = 10 sin θ
= 10 * sin 45°
= 6.69 m
considering vertical height
V (y) = v sin θ
= 5.66 sin 42
= 3.79 m/s
Next step : Calculate height ( h₂ )
v² = u² + 2 a s
0 = 3.79² - 2 * 9.8 * h₂
therefore : h₂ = 0.73 m
The maximum height attained by the rock ( H ) = h₁ + h₂
= 6.69 + 0.73 = 7.42 m
Hence we can conclude that The maximum height that the rock attains is : 7.42 m
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