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adell [148]
2 years ago
13

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp

ike is difficult. Suppose a ball is spiked from a height of 2.10 m with an initial speed of 21.0 m/s at a downward angle of 14.0°. How much farther on the opposite floor would it have landed if the downward angle were, instead, 6.0°?

Physics
1 answer:
MAVERICK [17]2 years ago
7 0

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

V_{Ax}=V_{A}cos\theta

V_{Ax}=(21m/s)cos(-14^{o})

V_{Ax}=20.38 m/s

V_{Ay}=V_{A}sin\theta

V_{Ay}=(21m/s)sin(-14^{o})

V_{Ay}=-5.08 m/s

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

t= -0.9159s    or   t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

x_{B}=V_{Bx}t

x_{B}=(20.88m/s)(0.468s)

x_{B}=9.77m

So once we got the two distances we can now find the difference between them:

x_{B}-x_{A}=9.77m-6.46m=3.31m

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

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Answer:

\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}

Explanation:

We need to find the acceleration of the 2 kilogram object. Let's complete this in 2 steps.

<h3>1. Force of 1st Object </h3>

First, we can find the force of the first, 8 kilogram object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

F=m \times a

The mass of the object is 8 kilograms and the acceleration is 1.6 meters per square second.

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Substitute these values into the formula.

F= 8 \ kg * 1.6 \ m/s^2

Multiply.

F= 12.8 \ kg*m/s^2

<h3>2. Acceleration of the 2nd Object </h3>

Now,  use the force we just calculated to complete the second part of the problem. We use the same formula:

F= m \times a

This time, we know the force is 12.8 kilograms meters per square second and the mass is 2 kilograms.

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\frac {12.8 \ kg*m/s^2}{2 \ kg}= \frac{2\ kg* a}{2 \ kg}

\frac {12.8 \ kg*m/s^2}{2 \ kg}=a

The units of kilograms cancel.

\frac {12.8}{2}\ m/s^2=a

6.4 \ m/s^2=a

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Answer:

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Explanation:

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Electron concentration n= 2.7\times10^{18}\ cm^{-3}

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