Question

Ann is at the neighborhood park playing with her dog, Biscuit. She throws a tennis ball straight up into the air from a height of 4 feet with a velocity of 40 feet per second. After the ball starts to come back down, Biscuit jumps up and catches the ball at a height of 5 feet. Which equation can you use to find how many seconds the ball is in the air before Biscuit catches it? If an object travels upward at a velocity of v feet per second from s feet above the ground, the object's height in feet, h, after t seconds can be modeled by the formula h=– 16t2+vt+s.

Answer 1

To determine how many seconds the ball is in the air before Biscuit catches it at 5 feet, use the equation h = –16t² + 40t + 4, where 'h' is the height after 't' seconds, 'v' is the initial velocity, and 's' is the initial height.

Step-by-step explanation:

• The specific scenario involves a ball being thrown vertically into the air and requires the use of a kinematic equation to solve for the time it takes for the ball to reach a certain height.
• To find the time 't' when the ball is caught by Biscuit at 5 feet, the provided kinematic equation h = –16t² + vt + s can be used, where 'h' is the height, 'v' is the initial velocity, and 's' is the initial height.
• Given that the initial velocity is 40 feet per second (v = 40) and the initial height is 4 feet (s = 4), and the ball is caught at 5 feet (h = 5), we can solve for 't' by plugging these values into the kinematic equation: 5 = –16t² + 40t + 4.
• The solutions to this equation 5 = –16t² + 40t + 4, will give us the time 't' at which Biscuit catches the ball.