Answer:
a) 7√3 m/s ≈ 12.124 m/s
b) 7 m/s
c) 2.5 m
d) 1 3/7 s ≈ 1.429 s
e) 10√3 m ≈ 17.32 m
Explanation:
a) It might help to draw a diagram. The horizontal component of a velocity vector inclined by angle α from the horizontal is cos(α) times the magnitude of the velocity. Here, that is ...
Vx = (14 m/s)(cos(30°)) = (14 m/s)((√3)/2) = 7√3 m/s
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b) The vertical component of the velocity is sin(α) times the magnitude of the velocity. Here, that is ...
Vy = (14 m/s)(sin(30°)) = (14 m/s)(1/2) = 7 m/s
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c) Energy considerations (kinetic to potential) tell you the maximum height d that will be reached when ...
(Vy)² = 2gd
d = (7 m/s)²/(2·9.8 m/s²) = 2.5 m
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d) We interpret the question to mean the time it takes for the ball to return to the ground from the time it left the ground (as opposed to the time it reached max height).
The vertical height can be computed from ...
h(t) = -1/2gt² +Vy·t
This will be zero when at the time when the ball is on the ground ...
0 = t((-g/2)t +Vy)
t = 0
t = 2Vy/g = 2(7 m/s)/(9.8 m/s²) = 10/7 s ≈ 1.429 s
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e) Multiplying the time in air by the horizontal speed gives the horizontal distance:
(10/7 s)(7√3 m/s) = 10√3 m
