when a ball is thrown up into the air it makes the shape of a parabola the equation s=-16t^2+v*t+k gives the height nof the ball at any time t in seconds where "v" is the initial velocity (speed)in ft/sec and "k" is the initial height in feet
What is the question? What are you required to find?
A ball is kicked upward with an initial velocity of 56 feet per second. The ball's height h (in feet) can be expressed as a function of time t (in seconds) by the equation h = -16t2 + 56t. How much time does the ball take to return to the ground?
To determine the height of a ball at any given time when thrown up into the air, we can use the equation s = -16t^2 + vt + k, where "s" represents the height in feet, "t" represents the time in seconds, "v" represents the initial velocity (speed) in feet per second, and "k" represents the initial height in feet.
Let's break down the equation to understand it better:
- The term -16t^2 represents the effect of gravity pulling the ball down. The negative sign indicates that the height decreases as time increases.
- The term vt represents the upward motion of the ball due to its initial velocity.
- The term k represents the initial height of the ball. If the ball was thrown from the ground (k = 0), this term does not affect the equation.
When using this equation to find the height of the ball at a specific time (t), you can substitute the known values of "v" and "k" into the equation. By solving for "s", you will obtain the corresponding height of the ball at that time t.
For example, if the initial velocity is 30 ft/sec (v = 30) and the ball was thrown from a height of 5 feet (k = 5), we can use the equation s = -16t^2 + 30t + 5 to find the height at any given time t.