Note the construction of the height equation in the problem above. The equation for the object's height s at time t seconds after launch is s( t) = −4.9 t 2 + 19.6 t + 58.8, where s is in meters. An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform.Yes, you'll need to keep track of all of this stuff when working with projectile motion. The projectile-motion equation is s( t) = −½ g x 2 + v 0 x + h 0, where g is the constant of gravity, v 0 is the initial velocity (that is, the velocity at time t = 0), and h 0 is the initial height of the object (that is, the height at of the object at t = 0, the time of release).

If a projectile-motion exercise is stated in terms of feet, miles, or some other Imperial unit, then use −32 for gravity if the units are meters, centimeters, or some other metric unit, then use −9.8 for gravity. And this duplicate 'per second' is how we get 'second squared'. So, if the velocity of an object is measured in feet per second, then that object's acceleration says how much that velocity changes per unit time that is, acceleration measures how much the feet per second changes per second. What does 'per second squared' mean?Īcceleration (being the change in speed, rather than the speed itself) is measured in terms of how much the velocity changes per unit time. The 'minus' signs reflect the fact that Earth's gravity pulls us, and the object in question, downward. The g stands for the constant of gravity (on Earth), which is −9.8 meters per second square (that is meters per second per second) in metric terms, or −32 feet per second squared in Imperial terms.

In projectile-motion exercises, the coefficient on the squared term is −½ g.

Freefall position function in meters