A projectile is fired at an angle
of 32 degrees above the horizontal with an initial speed of 44 meters per
second. What is the maximum upward vertical
displacement of the projectile from its launch position?
Okay, so, let’s say that this is
ground level and this orange line shows us the trajectory of this projectile. The projectile is fired at an
initial speed we’ll call 𝑣 sub 𝑖 and at an angle we’ll call 𝜃. We want to solve for the maximum
upward vertical displacement the projectile reaches, that is, this displacement
here. We’ll call it ℎ sub max.
Since we’re working with a
projectile and we know its initial speed and direction of travel, we can recall an
equation for the maximum height reached by a projectile in terms of its initial
speed and its direction. Since we’re given 𝑣 sub 𝑖 and 𝜃
and we also know that the acceleration due to gravity is 9.8 meters per second
squared, we can plug these values into our equation for ℎ sub max. When we enter this expression on
our calculator, to two significant figures, we find a result of 28 meters. This is the maximum vertically
upward displacement of our projectile.