a ball is thrown at an original speed of 8.0 m/s at an angle of 35 degrees above the horizontal. What is the speed of the ball when it returns to the same horizontal level?

To find the speed of the ball when it returns to the same horizontal level, we can break down the motion into horizontal and vertical components.

First, let's find the horizontal component of the initial velocity. The horizontal speed remains constant throughout the motion, so it will be the same as the initial horizontal component of the velocity.

Horizontal component of velocity = initial velocity * cos(angle)
= 8.0 m/s * cos(35°)
≈ 8.0 m/s * 0.819
≈ 6.552 m/s

Next, let's find the vertical component of the initial velocity. The vertical speed changes due to the acceleration of gravity.

Vertical component of velocity = initial velocity * sin(angle)
= 8.0 m/s * sin(35°)
≈ 8.0 m/s * 0.574
≈ 4.592 m/s

When the ball returns to the same horizontal level, the vertical component of velocity will have the same magnitude but opposite sign.

So, the speed of the ball when it returns to the same horizontal level can be calculated using the Pythagorean theorem.

Speed = √(horizontal component of velocity^2 + vertical component of velocity^2)
= √(6.552 m/s)^2 + (4.592 m/s)^2)
≈ √(42.92 + 21.12)
≈ √(64.04)
≈ 8.00 m/s

Therefore, the speed of the ball when it returns to the same horizontal level is approximately 8.00 m/s.