Answered

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A tennis ball is hit with an initial velocity of 15 m/s at an angle of
40°. What is the horizontal distance of the ball after 2 seconds?
(round to the nearest meter)

Sagot :

NusrathC

Answer:

Horizontal acceleration is 0 in a projectile motion

Explanation:

first we have to find the horizontal velocity

15cos40 = 11.49ms^-1

and then use the following equation to find the distance

[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = 11.49 \times 2 \\s = 22.98m[/tex]

View image NusrathC
Cricetus

The horizontal acceleration would be zero.

Given:

  • Initial velocity = 15 m/s
  • Angle = 40°

According to the question,

→ [tex]15 \ Cos 40^{\circ} = 11.49 \ m/s^{-1}[/tex]

By using the relation, we get

→ [tex]s = ut +\frac{1}{2}at^2[/tex]

By substituting the values, we get

     [tex]= 11.49\times 2[/tex]

     [tex]= 22.98 \ m[/tex]

Thus the response above is correct.

Learn more about distance here:

https://brainly.com/question/23397620

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