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pantera1 [17]
3 years ago
12

Suppose the designers of the water slide want to adjust the height h above the water so that riders land twice as far away from

the bottom of the slide. What would be the necessary height above the water?
A. 1.2 m
B. 1.8 m
C. 2.4 m
D. 3.0 m
Physics
1 answer:
Nitella [24]3 years ago
4 0

Answer:

option C

Explanation:

given,

Range is doubled

for range to be doubled height should become 4 times.

formula to calculate the horizontal range

          R = v T_{flight}

using equation of motion

s = u t + \dfrac{1}{2}at^2

t=\sqrt{\dfrac{2s}{g}}

now,

      R = v\sqrt{\dfrac{2h}{g}}

where, h is the height of the

Assuming the height be equal to 0.6 m

to calculate the new height

         h_{new}= 4 h

         h_{new}= 4\times 0.6

        h_{new}= 2.4\ m

hence, the correct answer is option C

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(a) 1.939 m/h

(b) 0.926 m/h

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Here, we have the water depth given by the function of time;

D(t) = 7 + 5·cos[0.503(t-6.75)]

Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;

D'(t) = \frac{d(7 + 5\cdot cos[0.503(t-6.75)])}{dt}

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(a) at 5:00 AM = 5 -  0:00 = 5

D'(5) =  -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h

(b) at 6:00 AM = 6 -  0:00 = 6

D'(5) =  -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h

(c) at 7:00 AM = 7 -  0:00 = 7

D'(5) =  -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h

(d) at Noon 12:00 PM = 12 -  0:00 = 12

D'(5) =  -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.

4 0
2 years ago
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