How To Calculate Height & Velocity

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.

Step 1

Write down this formula:

\(v_f=v_0+at\)

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.

Step 2

In the equation, v_f, 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.

Step 3

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:

\(t=\frac{v_f-v_0}{a}\)

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.

Step 4

Replace vf with zero to yield this simplified equation:

\(t=\frac{0-v_0}{a}=\frac{v_0}{a}\)

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).

Step 5

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

\(t=\frac{10}{a}\)

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.

Step 1

Write down this equation:

\(h=v_0t+\frac{1}{2}at^2\)

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.

Step 2

Plug the known values for t and v0 values as shown below:

\(h=10(0.31)+\frac{1}{2}32(0.31)^2\)

Step 3

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.