What equations should I use to find the Max height, time, and distance? I'm stuck on what equations to use

Question

A cannon tilted upward at  = 35.1  fires a cannonball with a speed of 91.4  on level ground.

What is the maximum height reached by the cannonball (above the ground)?

At what time after being fired does the cannonball reach it's maximum height above the ground?

What is the distance that the cannonball travels horizontally before reaching the ground again (assuming it was shot at exactly ground level with the velocity given)?

Answer 1

So there are three equations of motion:

(1) v = v₀ + at

(2) s = v₀t + (1/2)at²

(3) 2as = v² - v₀²

What is the maximum height reached by the cannonball (above the ground)?

You know the initial speed of the cannonball, to the tilted direction. If you're concerned about the maximum height, you need to extract the vertical component of this initial speed. That would be 91.4*sin(35.1). Let's call this vertical initial speed "v_0V".

You know the acceleration: It's 9.8 m/s², downward direction. If you set your v_0V to be positive, your acceleration "a" has a negative sign because it points the other way.

You also know that, if a cannonball reaches the maximum height, its vertical speed (v_V) must be 0.

So you use an equation that gives your maximum height (vertical displacement. Let's call this s_V) when you know a, v_V, and v_0V. Looks like you're using 2as = v² - v₀²

At what time after being fired does the cannonball reach it's maximum height above the ground?

Again, you know the acceleration a the initial vertical speed v_0V, and the fact that v_V = 0 at the maximum height. You want to use an equation that gives you the time when you know v_0V, v_V, and a. That would be v = v₀ + at.

What is the distance that the cannonball travels horizontally before reaching the ground again?

You figured out the time you're taking to reach the maximum height from the previous part. It will take exactly double amount of time for the cannonball to fall back from that maximum height. Anyways, we know how much time this is.

Now you need to extract the horizontal component of your initial speed, which is 91.4*cos(35.1). I'm calling this v_0H for being horizontal.

Well, we talked about how there is only vertical acceleration because of how gravity works. That means the horizontal acceleration is 0.

You want to figure out the horizontal displacement, knowing the horizontal initial speed, traveling time, and the fact that there's no acceleration. s = v₀t + (1/2)at² is the right one to use here. Because a = 0, you'll just need to do s = v₀t : multiply your horizontal speed with the traveling time.