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4 years ago

A ball is thrown horizontally from a cliff at 8m/s .In what direction is the ball moving 2s later?

Physics

1 answer:

horrorfan [7]4 years ago

5 0

Answer:

67.8 degrees below horizontal

Explanation:

In order to understand the direction of the ball after a certain time t, we need to find its horizontal and vertical components of the velocity.

- Horizontal component: the horizontal component of the velocity of the ball is constant, since there are no forces acting along this direction. Therefore, its horizontal velocity after t = 2 s is

vx = 8 m/s

- Vertical component: the vertical component of the velocity of the ball is given by

vy = u + at

where

u = 0 is the initial vertical velocity

a = g = -9.8 m/s^2 is the acceleration of gravity

Substituting t = 2s,

vy = 0 + (-9.8 m/s^2)(2s )=-19.6 m/s

where the negative sign indicates it points downward.

So, the direction of the ball (given as angle below the horizontal) is:

$\theta = tan^{-1} (\frac{|v_y|}{v_x})=tan^{-1} (\frac{19.6 m/s}{8 m/s})=67.8^{\circ}$

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3 years ago

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A golf ball is struck with a five iron on level ground. it lands 100.0 m away 4.60 s later. what was the magnitude and direction

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consider the motion in x-direction

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Using the equation

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consider the motion along y-direction

$v_{oy}$ = initial velocity in y-direction = ?

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$a_{y}$ = acceleration along x-direction = - 9.8 m/s²

t = time of travel = 4.60 sec

Using the equation

Y = $v_{oy}$ t + (0.5) $a_{y}$ t²

0 = $v_{oy}$ (4.60) + (0.5) (- 9.8) (4.60)²

$v_{oy}$ = 22.54 m/s

initial velocity is given as

$v_{o}$ = sqrt(( $v_{ox}$ )² + ( $v_{oy}$ )²)

$v_{o}$ = sqrt((21.7)² + (22.54)²) = 31.3 m/s

direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg

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4 years ago

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2 years ago

A ball is thrown horizontally from a cliff at 8m/s .In what direction is the ball moving 2s later?​

A ball is thrown horizontally from a cliff at 8m/s .In what direction is the ball moving 2s later?