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A ball is thrown straight upward and returns to the thrower’s hand after 2.96 s in the air. A second ball is thrown at an angle of 36.0 ◦ with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically

Respuesta :

audreyn06

Answer: 35.8 m/s

Explanation:

First, we need to find the height that the first ball reaches.

We can use the kinematics equation v = v₀ + at to solve for the initial velocity of the first ball, where:

  • v is the final velocity
  • v₀ is the initial velocity
  • a is the acceleration
  • t is the time

In this case:

  • v = 0 m/s since the ball is at rest when it is caught
  • We are trying to solve for v₀
  • a = -9.8 m/s² since the ball is in projectile motion, where it is only subject to gravity
  • t = 2.96 s

Let's plug these values in:

0 = v₀ + (-9.8)(2.96)

Solve for v₀:

v₀ = 29 m/s

Since we know the initial velocity of the ball, we can determine the maximum height it reached with the kinematics equation v² = v₀² + 2ad, where d is the displacement. In this case, the displacement is the maximum height reached. Let's plug our values into the equation:

0 = 29² + 2(-9.8)d

Solve for d:

d = 42.9 m

The first ball reached a maximum of 42.9 meters.

We can use the same kinematics equation v² = v₀² + 2ad to solve for the speed the second ball must be thrown at to reach the same height. We just need to solve for v₀:

0 = (cos(36)v₀)² + 2(-9.8)(42.9)

Note that there is a cos(36) being multiplied to v₀. This is because the second ball is being thrown at an angle of 36° with the horizontal.

Solve for v₀:

0 = cos²(36)v₀² - 840.84

840.84 = cos²(36)v₀²

Divide both sides by cos²(36):

v₀² = 1284.68920539

v₀ = 35.8 m/s

The second ball must be thrown at a speed of 35.8 m/s in order to reach the same height as the one thrown vertically.

Learn more about kinematics here: https://brainly.com/question/26407594

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