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LUCKY_DIMON [66]
1 year ago
12

Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2

0 ft/s.Step 1 of 2: Construct a set of parametric equations describing the shot. Round all final values to the nearest tenth.Answer

Physics
1 answer:
Korvikt [17]1 year ago
8 0

Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

x=v\cos\theta t

On substituting the known values,

\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}

The change in the height, y of the basketball is given by,

y=-v\sin\theta t+\frac{1}{2}gt^2

Where g is the acceleration due to gravity.

On substituting the known values,

\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}

Final answer:

The parametric equations describing the shot are

\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}

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3 years ago
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A video game includes an asteroid that is programmed to move in a straight line across a 17-inch monitor according to the equati
Ainat [17]

Answer:

The asteroid's acceleration at this point is 2.71\ m/s^2

Explanation:

The equation that governs the trajectory of asteroid is given by :

x=6.5t-2.3t^3

The velocity of asteroid is given by :

v=\dfrac{dx}{dt}\\\\v=\dfrac{d(6.5t-2.3t^3)}{dt}\\\\v=6.5-6.9t^2

At some point during the trip across the screen, the asteroid is at rest. It means, v = 0

So,

6.5-6.9t^2=0\\\\t=0.971\ s                      

Acceleration,

a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6.5-6.9t^2)}{dt}\\\\a=-13.8t                        

Put t = 0.971 s

a=-13.8\times 0.197\\\\a=-2.71\ m/s^2

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6 0
3 years ago
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Ghella [55]

Answer:

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Explanation:

bc ik it is :3

5 0
3 years ago
The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?
Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours.  So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes.  So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = <em>500 g</em> .

============================================

<u>Another way</u>:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = <em>500 g </em>.

5 0
3 years ago
65 POINTS! PLEASE ANSWER EVERY QUESTION! NEED HELP ASAP!
otez555 [7]
Maybe you can split up the questions. I will try to answer your first question.

1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)

2. Momentum: p = m₁v₁ + m₂v₂

m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B

Momentum before the collision:
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Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0

3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.

4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7

5.You figure out.



3 0
3 years ago
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