Answered

Experience the convenience of getting your questions answered at IDNLearn.com. Our community provides accurate and timely answers to help you understand and solve any issue.

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 :

NusrathCidn

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
Cricetusidn

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

We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For clear and precise answers, choose IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.