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finlep [7]
3 years ago
14

Two golf balls are hit from the same point on a flat field. Both are hit at an angle of 20° above the horizontal. Ball 2 has twi

ce the initial speed of ball 1. If ball 1 lands a distance d1 from the initial point, at what distance d2 does ball 2 land from the initial point? (Neglect any effects due to air resistance.)
A) dd_{2} = 2d_{_{1}}

B) d_{2} = 0.5d_{_{1}}

C)d_{2} = 8d_{_{1}}

D) d_{2} = d_{_{1}}

E) d_{2} = 4d_{_{1}}
Physics
1 answer:
Nutka1998 [239]3 years ago
5 0

Answer:

E) d_{2} = 4d_{_{1}}

Explanation:

We know that the range of a projectile is given by:

R=\frac{u^2.sin\ \2\theta}{g}

where:

u = initial velocity of the projectile

\theta = angle of projection of projectile

<u>Now, range for ball 1 :</u>

d_1=\frac{u^2.sin\ \2\theta}{g}

<u>Now, range for ball 2 with double velocity and all the other parameters being the same:</u>

d_2=\frac{(2u)^2.sin\ \2\theta}{g}

d_2=4\times \frac{u^2.sin\ \2\theta}{g}

d_2=4\times d_1

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Question is missing. Found on google:

a) What are the components of the acceleration of the fish?  

(b) What is the direction of its acceleration with respect to unit vector î?

(c) If the fish maintains constant acceleration, where is it at t = 30.0 s?

(a) (0.73, -0.47) m/s^2

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v_x = u_x + a_x t\\a_x = \frac{v_x-u_x}{t}=\frac{15.0-4.00}{15.0}=0.73 m/s^2

v_y = u_y + a_y t\\a_y = \frac{v_y-u_y}{t}=\frac{-6.00-1.00}{15.0}=-0.47 m/s^2

(b) -32.8^{\circ}

The direction of the acceleration vector with respect to i can be found by using the formula

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a_y is the vertical component of the acceleration

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a_x = 0.73 m/s^2

a_y = -0.47 m/s^2

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(c) r=(460.5 i - 185.1 j )m

The initial position of the fish is

r_0 = (12.0 i -3.60 j) m

The generic position r at time t is given by

r= r_0 + ut + \frac{1}{2}at^2

where

u=(4.00 i + 1.00 j) m/s is the initial velocity

a=(0.73 i -0.47 j) m/s^2 is the acceleration

Substituting t = 30.0 s, we find the final position of the fish. Separating each component:

r_x =12.0 + (4.00)(30) + \frac{1}{2}(0.73)(30)^2=460.5 m\\r_y = -3.60 + (1.00)(30) + \frac{1}{2}(-0.47)(30)^2=-185.1 m

So the final position is

r=(460.5 i - 185.1 j )m

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