# How to Find the Maximum Height at the Top of a Trajectory

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What goes up must come down. But you might want to know how far something goes up before it starts to come down. You'll need to know how fast it is moving when it starts, the angle of the trajectory, and the acceleration due to gravity of whatever planet you happen to be standing on.

• Determine the initial velocity and the angle at which the projectile is being launched. This will be somewhere between zero and 90 degrees. If it's zero, then you don't need any further calculation, because the maximum height is zero. The projectile isn't moving up at all. It's just moving sideways. If it's 90, it means that the object is going straight up. If it's somewhere in between, you'll need some elementary trigonometry to figure out the angle.

• Take the sine of the angle and multiply it by the initial velocity. The result is the vertical component of the initial velocity. If the angle is 90, the sine is equal to one, which means the entire initial velocity is the vertical component, as expected. Suppose instead that the object is launched at a 30-degree angle from the ground. The sine of 30 degrees is .5. If the initial velocity was 10 meters per second, the vertical component would be 5 meters per second.

• Use the equation for the object's vertical velocity as a function of time. v =v0 +at where v0 = the object's initial velocity and a = acceleration. Assuming this object is being projected on the earth, the acceleration would be equal to -9.8 meters/second^2. If the object is being projected somewhere else in the universe, just plug in the local acceleration due to gravity. The object will reach its maximum height when velocity equals zero, so, using the example above, the equation becomes: 0 = (5 meters / second) -(9.8 meters / second^2) * t. Solving for t: t = (5 meters / second) / (9.8 meters / second^2) = .51 seconds. The object will stop moving up and start to move down at .51 seconds.

• Use the equation for the object's vertical position y= v0 t +1/2 at^2 where v0 = the initial vertical velocity and t = time. Plug in the values you know. In the example, that yields the equation: y = 5 meters / second .51 seconds - 4.9 meters / second^2 * (.51 seconds)^2 = 1.27 meters. The object will reach a maximum height of 1.27 meters at the top of its trajectory.

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