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During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial velocity v0 = 26 m/s at an angle θ = 17° above horizontal. Let the origin of the Cartesian coordinate system be the ballʼs position at impact. Air resistance may be ignored throughout this problem.


a) express the magnitude of the ball's initial horizontal velocity, v0x, in terms of v0 and theta.


b) express the magnitude of the ball's inital vertical velocity, v0y, in terms of v0 and theta.


c) find the ball's maximum vertical height, hmax, in meters above ground.


d) create an expression in terms of v0, theta, and g for the time (tmax) it takes the ball to travel to its maximum vertical height.


e) calculate the horizontal distance, xmax, in meters the ball has traveled when it returns to ground level.

Respuesta :

Answer: a) vox = vo × cos θ, b) voy =vo× sin θ,

c) H=2.94 m, d) t = vo sinθ / g, e) R = 38.57 m

Explanation:

A)

The velocity v0 is at angle θ to the horizontal.

The horizontal component of vo (vox), vo and the vertical component of vo (voy) all form a right angle triangle.

With vo as the hypotenus, vox as the adjacent and voy as the opposite.

To get vox, we relate vo and vox ( hypotenus and adjacent)

From trigonometry

Cos θ relates hypotenus and adjacent, hence we have that

Cos θ = vox/vo

vox = vo × cos θ

B)

To get the vertical component of vo, we relate vo and voy ( hypotenus and opposite).

According to trigonometry, sin θ relates hypotenus and opposite, hence we have that

Sin θ = voy/vo

voy =vo× sin θ

C)

The formulae for the maximum height of a projectile motion is given as

H = vo² (sin θ)²/2g

Where g = acceleration due to gravity = 9.8 m/s²

By substituting the parameters, we have that

H = 26² × (sin 17)²/2(9.8)

H = 676 × 0.0854/19.6

H = 57.7304/ 19.6

H = 2.94 m

D)

This is the motion of a projectile and the conditions at maximum height are vy = 0 and ay = - g

From the equation of motion

vy = voy - gt

0 = voy - gt

But voy = vo sinθ

0 = vo sinθ - gt

gt = vo sinθ

t = vo sinθ / g

E)

The horizontal distance covered formulae is given by

R = u² sin2θ/g

R = 26² × sin 2(17)/9.8

R = 676 × sin 34/ 9.8

R = 378.014/ 9.8

R = 38.57 m

The correct Answer is:

  • a) vox = vo × cos θ, b) voy =vo× sin θ,
  • c)  H=2.94 m, d) t = vo sinθ / g, e) R = 38.57 m

A) When The velocity v0 is at angle θ to the horizontal.

  • When The horizontal component of vo (vox), vo, and also that the vertical component of vo (voy) all form a right angle triangle.
  • Although when With vo as the hypotenuse, vox as the adjacent and voy as the opposite.
  • Then To get vox, we relate vo and also that vox ( hypotenuse and adjacent)
  • When From trigonometry
  • Also, Cos θ relates hypotenuse and adjacent, hence we have that
  • Then Cos θ = vox/vo
  • Then vox = vo × cos θ

B) When To get the vertical component of vo, we relate to and also voy ( hypotenuse and also opposite).

  • According to trigonometry, sin θ relates to the hypotenuse and also that opposite, hence we have that
  • Then Sin θ = voy/vo
  • Then voy =vo× sin θ

C) When The formulae for the maximum height of a projectile motion is given as

  • H = vo² (sin θ)²/2g
  • Where that g = acceleration due to gravity = 9.8 m/s²
  • By substituting the parameters, we have that
  • Then H = 26² × (sin 17)²/2(9.8)
  • Then H = 676 × 0.0854/19.6
  • Then H = 57.7304/ 19.6
  • Then H = 2.94 m

D) When This is the motion of a projectile and the conditions at maximum height are vy = 0 and ay = - g

  • From the equation of motion
  • vy = voy - gt
  • 0 = voy - gt
  • But voy = vo sinθ
  • 0 = vo sinθ - gt
  • gt = vo sinθ
  • t = vo sinθ / g

E) When The horizontal distance covered formulae is given by

  • R = u² sin2θ/g
  • R = 26² × sin 2(17)/9.8
  • R = 676 × sin 34/ 9.8
  • R = 378.014/ 9.8
  • R = 38.57 m

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