A wheel which is initially at rest starts to turn with a constant angular acceleration. after 4 seconds it has made 4 complete r
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Answer:
(a) 3.22 m
(b) The vertical velocity, , at maximum height is 0 m/s, the horizontal velocity, vₓ, is 12.72 m/s
(c) The acceleration at maximum height = g = 9.81 m/s²
(d) The time it takes for the paper to reach the balcony is 1.212 seconds
(e) The horizontal range, of the paper is 15.42 m.
Explanation:
(a) Given that we re given a projectile motion, we have the following governing equations;
y = y₀ + v₀·sin(θ₀)·t - 0.5×g·t²
= v₀·sin(θ₀) - g·t
Where:
y = Height of the paper
y₀ = Initial height of the paper = Ground level = 0
v₀ = Inititial velocity of the paper = 15.0 m/s
θ₀ = Angle in which the paper is thrown = 32° above the horizontal
g = Acceleration due to gravity = 9.81 m/s²
t = Time taken to reach the height h
= Vertical velocity of the paper
At maximum height, = 0, therefore;
= v₀·sin(θ₀)·t - g·t = 0
v₀·sin(θ₀) = g·t
t = v₀·sin(θ₀)/g = 15×sin(32°)/9.81 = 0.81 seconds
y = y₀ + v₀·sin(θ₀)·t - 0.5×g·t² = 0 + 15×sin(32°)×0.81-0.5×9.81×0.81² = 3.22 m
(b) The vertical velocity, , at maximum height = 0 m/s, the horizontal velocity, vₓ, = 15×cos(32°) = 12.72 m/s
(c) The acceleration at maximum height = g = 9.81 m/s²
(d) The time it takes to maximum height = 0.81 seconds
The time the paper will take to fall to 1.25 m above the ground, which is 3.22 - 1.25 = 1.97 meters below maximum height is therefore given as follows;
y = y₀ + v₀·sin(θ₀)·t - 0.5×g·t²
Where:
v₀ = 0 m/s at maximum height
y = -1.97 m downward motion
y₀ = 0 starting from maximum height downwards
1.97 = 0 + 0·sin(θ₀)·t - 0.5×9.81×t²
-1.97 = - 0.5×9.81×t²
t = (-1.97)/(-0.5*9.81) = 0.402 seconds
The time the paper will take to fall to 1.25 m above the ground = 0.81+0.402 = 1.212 seconds
Therefore, the time it takes for the paper to reach the balcony = 1.212 seconds
(e) The horizontal range, x, is given by the relation;
x = x₀ + v₀·cos(θ₀)·
x₀ = Starting point of throwing the paper = 0
= Total time of flight of the paper
∴ x = x₀ + v₀·cos(θ₀)· = 0 + 15×cos(32°)×1.212 = 15.42 m
The horizontal range, of the paper = 15.42 m.
