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trapecia [35]
3 years ago
14

A ball is dropped from a height of 12 feet and returns to a height that is one-half of the height from which it fell. How far wi

ll the ball have traveled when it hits the ground for the fourth time?. . A. 12 feet. . B. 24 feet. . C. 3 feet. . D. 33 feet¥
Physics
2 answers:
almond37 [142]3 years ago
7 0

Answer: (D) 33 feet

First consider that the height is 12 feet and the ball is dropped from that height and the ball bounces back up to half the height that is 6 feet, then second time when the ball is dropped, according to the given question, the height becomes 6 feet and when the ball returns back then it returns back to half of the height that is 3 feet, following the similar path, we can write bounces one at a time:

1.  Ball is dropped from 12 feet and bounces back to 6 feet.

2. Ball is dropped from 6 feet and bounces back to 3 feet.

3. Ball is dropped from 3 feet and bounces back to 1.5 feet.

4. Ball is dropped from 1.5 feet and reaches the ground.

Adding the total distance traveled,we get

12+6+6+3+3+1.5+1.5=33 feet.

Liula [17]3 years ago
6 0
The best thing to do in order to calculate the distance of the ball taht would have traveled when it hits the ground for the fourth time is to list the height everytime it bounces. We calculate as follows:

<span>12+6+6+3+3+1.5+1.5 = 33 feet</span>
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A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police
Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. x=x_{0}+vt

2. x=x_{0}+v_{0}t+\frac{at^{2}}{2}

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

vt = v_{0}t+\frac{at^2}{2}\\\\   16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}

We simplify the fraction present and rearrange for our formula so that it equals 0:

0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0

In the very last step we factored a common factor t. There is two possible solutions to the equation at t=0 and:

0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\  0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at t=0 s (when the SUV passed the police car) and t=15.56s(when the police car catches up to the SUV)

8 0
3 years ago
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