The object's trajectory is described by a quadratic formula, and the height of the object after a specific time can be found using this formula. By applying the quadratic formula to this projectile motion scenario, we conclude that it will spend 3.79 seconds in the air before returning to the initial height of 10 feet.
The question relates to projectile motion, where an object is thrown upward and its height at various times is described by a quadratic equation. Given the initial upwards velocity of 77 feet per second and the starting height of 10 feet, the height h of the object after t seconds is given by the equation h = -16t^2 + 77t + 10. To determine the time at which the object reaches a certain height, one would typically use the quadratic formula to find t.
Example Calculation
When the object is thrown, it will reach the initial height of 10 feet twice, once on the way up and once on the way down. The quadratic formula might yield two times, and in this scenario, it yields t = 3.79 s and t = 0.54 s. We take the longer solution for the time the ball takes to return to the initial height because the ball will be in the air for that entire duration after being thrown upwards.
The time for projectile motion is determined completely by the vertical motion. Therefore, for an object with an initial vertical velocity of 77 feet per second (converted to meters per second if required) that lands 10 feet higher than its starting altitude, it will spend 3.79 seconds in the air.