Answer: 1.77 s
Explanation: In order to solve this problem we have to use the kinematic equation for the position, so we have:
xf= xo+vo*t+(g*t^2)/2 we can consider the origin on the top so the xo=0 and xf=29 m; then
(g*t^2)/2+vo*t-xf=0 vo is the initail velocity, vo=7.65 m/s
then by solving the quadratric equation in t
t=1.77 s
Answer:
d = 11.1 m
Explanation:
Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

Plugging this into the energy conservation equation and cancelling m, we get

Solving for d,

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Explanation:
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Answer:
(a) 17.37 rad/s^2
(b) 12479
Explanation:
t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0
w = v / r = 99 / 0.06 = 1650 rad/s
(a) Use first equation of motion for rotational motion
w = w0 + α t
1650 = 0 + α x 95
α = 17.37 rad/s^2
(b) Let θ be the angular displacement
Use third equation of motion for rotational motion
w^2 = w0^2 + 2 α θ
1650^2 = 0 + 2 x 17.37 x θ
θ = 78367.87 rad
number of revolutions, n = θ / 2 π
n = 78367.87 / ( 2 x 3.14)
n = 12478.9 ≈ 12479