Answered

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A ball is kicked off the ground at an angle of 20 at a speed of 24.67 m/s. a) Find the peak height of the ball. b.) Determine the flight time before it hit the ground. c.) Determine the horizontal distance.

Sagot :

Quinzenidn

PROJECTILE

A ball is kicked off the ground at an angle of 20° and at a speed of 24.67 m/s.

Given:

  • v¡ = 24.67 m/s
  • θ = 20°
  • g = 9.8 m/s²

a) How high will it go?

[tex]d_{y} = \displaystyle \frac{(v_{i} \sin \theta)^{2}}{2g}[/tex]

[tex]d_{y} = \displaystyle \frac{(24.67 \: m/s \times \sin 20^{\circ})^{2}}{2(9.8 m/s^{2})}[/tex]

[tex]d_{y} = \displaystyle \tt 3.63 \: m[/tex]

Answer: [tex] \large \sf \bold{ \blue{3.63 \: m}}[/tex]

b) How long will it take the ball before it strikes the ground?

[tex]t = \displaystyle \frac{2v_{i} \sin \theta}{g}[/tex]

[tex]t = \displaystyle \frac{2(24.67 \: m/s) \sin 20^{\circ}}{9.8 m/s^{2}}[/tex]

[tex]t = \displaystyle \tt 1.72 \: s[/tex]

Answer: [tex] \large \sf \bold{ \blue{1.72 \: s}}[/tex]

c) At what distance does the ball covers?

[tex]R = \displaystyle \frac{v_{i}\:^{2} \sin2 \theta}{g}[/tex]

[tex]R = \displaystyle \frac{(24.67 \: m/s)^{2} \sin 40^{\circ}}{9.8 m/s^{2}}[/tex]

[tex]R = \displaystyle \tt 39.92 \: m[/tex]

Answer: [tex] \large \sf \bold{ \blue{39.92 \: m}}[/tex]

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