Answer:
s(t) = –4.9t2 + 39.2t
Explanation:
What is the height (above ground level) when the object smacks into the ground? Well, zero, obviously. So I'm looking for the time when the height is s = 0. I'll set s equal to zero, and solve:
0 = –4.9t2 + 19.6t + 58.8
0 = t2 – 4t – 12
0 = (t – 6)(t + 2)
Then t = 6 or t = –2. The second solution is from two seconds before launch, which doesn't make sense in this context. (It makes sense on the graph, because the line crosses the x-axis at –2, but negative time won't work in this word problem.) So "t = –2" is an extraneous solution, and I'll ignore it.
The object strikes the ground six seconds after launch.
Note the construction of the height equation in the problem above. The initial launch height was 58.8 meters, and the constant term was "58.8". The initial velocity (launch speed) was 19.6 m/s, and the coefficient on the linear term was "19.6". This is always true for these up/down projectile motion problems. (If you have an exercise with sideways motion, the equation will have a different form, but they'll always give you that equation.) The initial velocity is the coefficient for the middle term, and the initial height is the constant term. And the coefficient on the leading term comes from the force of gravity. This coefficient is negative, since gravity pulls downward, and the value will either be "4.9" (if your units are "meters") or "16" (if your units are "feet"). In general, the format is:
s(t) = –gt2 + v0t + h0
...where "g" here is the "4.9" or the "16" derived from the value of the force of gravity (technically, it's half of the force of gravity, but you probably don't need to know that right now), "v0" ("vee-naught", or "vee-sub-zero") is the initial velocity, and "h0" ("aitch-naught", or "aitch-sub-zero") is the initial height.
Memorize this equation (or at least its meaning), because you may need to know this on the test.
An object in launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. What will be the object's maximum height? When will it attain this height?
Hmm... They didn't give me the equation this time. But that's okay, because I can create the equation from the information that they did give me. The initial height is 80 feet above ground and the initial speed is 64 ft/s. Since my units are "feet", then the number for gravity will be 16, and my equation is:
s(t) = –16t2 + 64t + 80
They want me to find the maximum height. For a negative quadratic like this, the maximum will be at the vertex of the upside-down parabola. So they really want me to find the vertex. From graphing, I know how to find the vertex; in this case, the vertex is at (2, 144):
h = –b/2a = –(64)/2(–16) = –64/–32 = 2
k = s(2) = –16(2)2 + 64(2) + 80 = –16(4) + 128 + 80 = 208 – 64 = 144
But what does this vertex tell me? According to my equation, I'm plugging in time values and extracting height values, so the input "2" must be the time and the output "144" must be the height. Copyright © Elizabeth Stapel 2004-2011 All Rights Reserved
It takes two seconds to reach the maximum height of 144 feet.
An object is launched from ground level directly upward at 39.2 m/s. For how long is the object at or above a height of 34.3 meters?
My units this time are "meters", so the gravity number will be "4.9". Since the object started at ground level, the initial height was 0. Then my equation is:
s(t) = –4.9t2 + 39.2t