Projectile motion problems are common on physics examinations. A projectile is an object that moves from one point to another along a path. Someone can toss an object into the air or launch a missile that travels in a parabolic path to its destination. A projectile's motion can be described in terms of velocity, time and height. If the values for any two of these factors are known, it is possible to determine the third.

## Solve for Time

Write down this formula:

Final Velocity = Initial Velocity + (Acceleration Due To Gravity * Time)

This states that the final velocity that a projectile reaches equals its initial velocity value plus the product of the acceleration due to gravity and the time the object is in motion. The acceleration due to gravity is a universal constant. Its value is approximately 32 feet (9.8 meters) per second. That describes how fast an object accelerates per second if dropped from a height in a vacuum. "Time" is the amount of time that the projectile is in flight.

Simplify the formula using short symbols as shown below:

vf = v0 + a * t

Vf, v0 and t stand for Final Velocity, Initial Velocity and Time. The letter “a” is short for “Acceleration Due To Gravity.” Shortening long terms makes it easier to work with these equations.

Solve this equation for t by isolating it on one side of the equation shown in the previous step. The resulting equation reads as follows:

t = (vf –v0) ÷ a

Since the vertical velocity is zero when a projectile reaches its maximum altitude (an object thrown upward always reaches zero velocity at the peak of its trajectory), the value for vf is zero.

Replace vf with zero to yield this simplified equation:

t = (0 – v0) ÷ a

Reduce that to get t=v0 ÷ a. This states that when you toss or shoot a projectile straight up into the air, you can determine how long it takes for the projectile to reach its maximum height when you know its initial velocity (v0).

Solve this equation assuming that the initial velocity, or v0, is 10 feet per second as shown below:

t = 10 ÷ a

Since a = 32 feet per second squared, the equation becomes t = 10/32. In this example, you discover that it takes 0.31 seconds for a projectile to reach its maximum height when its initial velocity is 10 feet per second. The value of t is 0.31.

## Solve for Height

You can use these same formulas to calculate a projectile’s initial velocity if you know the height it reaches when tossed into the air and the number of seconds that it takes to reach that height. Simply plug those known values into the equations and solve for v0 instead of h.

Write down this equation:

h = (v0 * t) + (a * (t*t) ÷ 2)

This states that a projectile’s height (h) is equal to the sum of two products -- its initial velocity and the time it is in the air, and the acceleration constant and half of the time squared.

Plug the known values for t and v0 values as shown below: h = (10 * 0.31) + (32 * (10 * 10) ÷ 2)

Solve the equation for h. The value is 1,603 feet. A projectile tossed with an initial velocity of 10 feet per second reaches a height of 1,603 feet in 0.31 seconds.