A juggler throws the balls 2.4 m in the air at an angle of 80° relative to the horizontal. What is the time between throws?

To determine the time between throws for a juggler throwing a ball to a height of 2.4 m at an angle of 80° to the horizontal, we can use kinematic equations. First, we need to find the time it takes for the ball to reach its peak height, and then double that time to account for the entire round trip of the ball.

1. Calculate the vertical component of velocity: The height ( h ) reached by the ball can be expressed as:

[

h = \frac{v_y^2}{2g}

]

where ( v_y ) is the initial vertical velocity and ( g ) is the acceleration due to gravity (( 9.81 , \text{m/s}^2 )). We can solve for ( v_y ):

[

v_y = \sqrt{2gh} = \sqrt{2 \times 9.81 , \text{m/s}^2 \times 2.4 , \text{m}} \approx \sqrt{47.088} \approx 6.86 , \text{m/s}

]

2. **Calculate the