Example 1 – A ball is thrown straight up from the top of a 128 foot tall building with an initial speed of 32 feet per second. The height of the ball as a function of time can be modeled by the function h(t) = –16t2 + 32t + 128.

So, “H/S = h/s.” For example, if s=1 meter, h=0.5 meter and S=20 meters, then H=10 meters, the height of the object.

Choosing Appropriate Units of Measure

h = v 0 y 2 2 g . This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical component of the initial velocity.

How to find the maximum height of a projectile?

1. if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest.
2. if α = 45°, then the equation may be written as:
3. if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that’s the case of horizontal projectile motion.

In mathematical terms, it can be considered as: If and , then . That means it takes 3.058 s for the ball to reach its maximum height.

1 Expert Answer So the highest point on the parabola is when x=3, or in our case t=3. So the ball reaches its maximum height after 3 seconds.

1 Expert Answer So the rocket will reach its maximum height after 4 seconds.