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Sholpan [36]
2 years ago
14

In the popular game "Angry Birds" (refer Figure), for the red bird to hit the green bird with their range 55.0 m, what should be

the launch angle (with respect to the horizontal) of the slingshot if the time of flight is 2.50 s? ​

Physics
1 answer:
Inga [223]2 years ago
4 0

Answer:

Explanation:

What are we to use as gravity in this imaginary world? I will ASSUME 9.81 m/s² and assume no air resistance.

horizontal velocity component

vx = 55.0/2.50 = 22 m/s

vertical initial velocity component

(12.0 - 10.0) = 0 + vy₀(2.50) + ½(-9.81)2.50²

vy₀ = 13.0625 m/s

θ = arctan(vy₀/vx) = arctan(13.0625/22) = 30.6997225... = 30.7°

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A funnel cloud is a funnel-shaped cloud of condensed water droplets. They usually appear with a rotating column of air. These extend from the bottom of a cloud that does not touch the ground or a water surface.
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2 years ago
Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom
Marianna [84]

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

   final velocity of Tarzan = v_f

law of conservation of energy

  PE_i + KE_i = PE_f + KE_f

mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2

       mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2

          mgh_i = \dfrac{1}{2}mv_f^2

             v_f = \sqrt{2gh_i}

                   = \sqrt{2gL(1- cos\theta)}

                   = \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}

                          = 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the  = 3 m/s

mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2

       mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2

          mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2

          gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2

             v_f = \sqrt{v_1^2+2gh_i}

             v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}

                       v_f= 11.29 m/s

3 0
3 years ago
a shell fired from a cannon at 60 ° from horizontal strikes a target 20m high at a distance 80m. Calculate the initial velocity
stich3 [128]

consider the motion along the X-direction

X = horizontal displacement = 80 m

V_{ox} = initial velocity along the x-direction = v Cos60

t = time of travel

using the equation

X = V_{ox}   t

80 = (v Cos60) (t)

t = 160/v                                         eq-1


consider the motion in vertical direction :

Y = vertical displacement = 20 m

V_{oy}  = initial velocity in Y-direction = v Sin60

a = acceleration = - 9.8 m/s²

t = time of travel = 160/v

using the equation

Y = V_{oy}  t + (0.5) a t²

20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²

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4 0
3 years ago
Hello I need help its due today!
telo118 [61]

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Explanation:

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2 years ago
What is the difference between a free body diagram and a vector diagram?​
Murrr4er [49]

Answer:

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Explanation:

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1 year ago
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