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mojhsa [17]
3 years ago
7

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

ssistant standing on the level ground 30.0 m downrange from the launch point throws an apple straight up with the minimum initial speed necessary to meet the path of the arrow. What is the initial speed of the apple and at what time after the arrow is launched should the apple be thrown so that the arrow hits the apple?
Physics
1 answer:
svlad2 [7]3 years ago
8 0

Answer:

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

the apple should be tossed after \Delta t=0.0173\ s

Explanation:

Given:

  • velocity of arrow in projectile, v=30\ m.s^{-1}
  • angle of projectile from the horizontal, \theta=20^{\circ}
  • distance of the point of tossing up of an apple, d=30\ m

<u>Now the horizontal component of velocity:</u>

v_x=v\ cos\ \theta

v_x=30\times cos\ 20^{\circ}

v_x=28.191\ m.s^{-1}

<u>The vertical component of the velocity:</u>

v_y=v.sin\ \theta

v_y=30\times sin\ 20^{\circ}

v_y=10.261\ m.s^{-1}

<u>Time taken by the projectile to travel the distance of 30 m:</u>

t=\frac{d}{v_x}

t=\frac{30}{28.191}

t=1.0642\ s

<u>Vertical position of the projectile at this time:</u>

h=v_y.t-\frac{1}{2}g.t^2

h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2

h=5.3701\ m

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

v'^2=u'^2-2g.h

0^2=u'^2-2\times 9.8\times 5.3701

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

v'=u'-g.t'

0=10.259-9.8\times t'

t'=1.0469\ s

<u>We observe that </u>t>t'<u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

\Delta t=t-t'

\Delta t=1.0642-1.0469

\Delta t=0.0173\ s

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Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(
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Answer

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     f = \dfrac{\omega}{2\pi}

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        k= \dfrac{2\pi}{\lambda}

        \lambda= \dfrac{2\pi}{k}

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e) direction of the motion will be in negative x-direction

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(Repost) I need help with this physics question. Thanks in advance! Answer ASAP.
Kitty [74]

The time taken for the tiny saliva to travel is 0.55 second.

The  horizontal distance traveled at speed of 4 m/s is 2.2 m.

The horizontal distance traveled at speed of 20 m/s is 11 m.

<h3>Time of motion of the tiny saliva</h3>

The time taken for the tiny saliva to travel is calculated as follows;

h = vt + ¹/₂gt²

where;

  • v is initial vertical velocity  = 0
  • g is the acceleration due to gravity

h = 0 + ¹/₂gt²

h = ¹/₂gt²

2h = gt²

t² = 2h/g

t = √(2h/g)

Substitute the given parameters and solve for time of motion;

t = √(2 x 1.5 / 10)

t = 0.55 second

<h3>Horizontal distance traveled at speed of 4 m/s</h3>

X = Vx(t)

X = (4 m/s)(0.55)

X = 2.2 m

<h3>Horizontal distance traveled at speed of 20 m/s</h3>

X = (20)(0.55)

X = 11 m

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

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