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Dock diving is a great form of athletic competition for dogs of all shapes and sizes. Sheba, the American Pit Bull Terrier, runs and jumps o
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Dock diving is a great form of athletic competition for dogs of all shapes and sizes. Sheba, the American Pit Bull Terrier, runs and jumps off the dock with an initial speed of 8.62 m/s at an angle of 28° with respect to the surface of the water. (Assume that the +x axis is in the direction of the run and the +y axis is up.) (a) If Sheba begins at a height of 0.85 m above the surface of the water, through what horizontal distance does she travel before hitting the surface of the water? m
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Physics
5 months
2021-09-02T12:57:01+00:00
2021-09-02T12:57:01+00:00 2 Answers
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Answers ( )
Answer:
3.139 m
Explanation:
Velocity = 8.62 m/s
Angle of flight = 28°
Height above water = 0.85 m
we are to take the +ve x-axis in the direction of run, and +ve y-axis up.
This is a projectile situation, and the horizontal distance of travel (range) is calculated as
R = (
sin2∅)/2g
where g is the acceleration due to gravity = +9.81 m/
sine it acts downwards
R = (
sin (2×28°))/(2 x 9.81)
R = (74.30 x 0.8290)/19.62
R = 61.60/19.62 = 3.139 m
Answer:
62.4m
Explanation:
a) Horizontal distance traveled = x =
* t
where,
and t = time in the air)
Time in the air t can be solved using the equation for y:
where,
= initial height i.e 0.85 m
and g = 9.8 m/s^2
When y = 0, the dog has hit the water.
So set y = 0 and solve for t.
0 = 0.85 m + (4.04 m/s)t – 0.5gt²
0.5t²-4.04t-0.85
Solve this quadratic formula for t: the solutions are t = 8.2 s and t = -0.2. Reject the negative solution, so t = 8.2 s.
How far does the dog travel horizontally in 9.2 s?
x =
* t = (8.62 m/s)
(cos 28)
8.2 s = 62.4 m