A ball thrown straight upward returns to its original level in 2.50s. A second ball is thrown at an angle of 36.degrees above the horizontal. What is the initial speed of the second ball if it also returns to its original level in 2.50
To find the initial speed of the second ball, we can use the principle of projectile motion and the fact that the ball returns to its original level. In projectile motion, the horizontal and vertical components of motion are independent of each other.
Let's break down the motion of the second ball into its horizontal and vertical components:
1. Horizontal Component: The horizontal component of velocity remains constant throughout the motion. Therefore, the horizontal distance traveled by the ball is the same as the horizontal component of its initial velocity times the time of flight.
2. Vertical Component: The vertical motion of the ball can be treated like an object thrown vertically upwards. The time taken for the ball to reach its highest point is half of the total time of flight.
Now, let's calculate the initial speed of the second ball:
Step 1: Calculate the horizontal distance traveled by the ball.
- The horizontal component of velocity is given by Vx = V * cos(θ), where V is the initial speed and θ is the angle above the horizontal.
- The horizontal distance traveled is given by Dx = Vx * t, where t is the total time of flight.
- Since the ball returns to its original level, the horizontal distance traveled is zero.
- Therefore, Dx = 0. So, Vx * t = 0.
- Since t is not equal to zero, Vx must be zero. Consequently, cos(θ) = 0.
Step 2: Determine the angle above the horizontal when cos(θ) = 0.
- For cos(θ) = 0 to hold true, θ must be equal to 90 degrees.
- In this case, the ball is thrown vertically upwards.
Step 3: Calculate the initial vertical component of velocity.
- The total time of flight (t) is given as 2.50s.
- The time taken to reach the maximum height is half of the total time of flight, so t/2 = 1.25s.
- The final vertical displacement during this time is 0 (since the ball returns to its original level).
- The equation for vertical displacement in projectile motion is: Dy = Vy0 * t - (1/2) * g * t^2, where Dy is the vertical displacement, Vy0 is the initial vertical component of velocity, and g is the acceleration due to gravity.
- Since the ball returns to its original level, Dy is equal to zero.
- Applying the equation: 0 = Vy0 * (t/2) - (1/2) * g * (t/2)^2.
- We can substitute the values t = 1.25s and g = 9.8 m/s^2 (approximate value on the surface of the Earth).
- Solving the equation for Vy0: Vy0 = (1/2) * g * (t/2).
- Vy0 = (1/2) * 9.8 * 1.25.
- Vy0 = 6.125 m/s.
Step 4: Calculate the initial speed of the second ball.
- The initial speed of the second ball is the magnitude of its initial velocity.
- Since the ball is thrown vertically upwards, its initial velocity is given by Vy0.
- Therefore, the initial speed of the second ball is 6.125 m/s.
So, the initial speed of the second ball thrown at an angle of 36 degrees above the horizontal is approximately 6.125 m/s.