Answer:
a = 2.72 ms⁻²
32.83 s
Explanation:
By using the kinematic equations you get,
v² = u² +2as and v = u + at where all terms in usual meaning
Using 1st equation,
89.3² = 0² + 2a×1465 ⇒ a = 2.72 ms⁻²
By 2nd equation,
89.3 = 0 + 2.72×t ⇒ t = 32.83 s
Define
u = 16 m/s, the vertical launch velocity
g = acceleration due to gravity, measured positive downward
s = vertical distance traveled
t = 21.2 s, total time of travel.
The vertical motion obeys the equation
s = ut - (1/2)gt²
When the rock is at ground level, s = 0.
Therefore
(16 m/s)(21.2 s) - 0.5*(g m/s²)*(21.2 s)² = 0
339.2 - 224.72g = 0
g = 1.5094 m/s²
Answer:
The acceleration due to gravity is 1.509 m/s² measured positive downward.
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Answer:
Taking forces along the plane
F cos θ - M g sin θ -100 = M a net of forces along the plane
F = (M a + M g * .5 + 100) / .866 solving for F
F = (80 * 1.5 + 80 * 9.8 * .5 + 100) / .866 = 707 N
F = 707 N acting along the plane
Fn = F sin θ + M g cos θ forces acting perpendicular to plane
Fn = 707 * 1/2 + 80 * 9.8 * .866 = 1030 Newtons forces normal to plane
(this would give a coefficient of friction of 100 / 1030 = .097 = Fn)