A toy rocket is shot upward into the air from a height of 1/2 meter above the ground with an initial velocity of 19.6 meters per second. The formula for the vertical motion of an object is h(t) = 1/2 at^2 + v t + s, where the gravitational constant a, is -9.8 meters per second squared, v is the initial velocity, s is the initial height, and h(t) is the meters modeled as a function of time, t.
Due to a malfunction, the toy rocket explodes when it reaches its maximum height. How high above the ground is the toy rocket when it explodes? Round to the nearest tenth.
To find the maximum height reached by the toy rocket before it explodes, we need to find the time at which the rocket reaches its highest point.
The rocket reaches its highest point when its vertical velocity is 0. We can find this time by setting the derivative of the height function equal to 0 and solving for t:
h'(t) = -9.8t + 19.6
0 = -9.8t + 19.6
9.8t = 19.6
t = 2
Now that we know the time at which the rocket reaches its highest point, we can plug this time back into the height function to find the maximum height:
h(2) = 1/2(-9.8)(2)^2 + 19.6(2) + 0.5
h(2) = -19.6 + 39.2 + 0.5
h(2) = 19.1
Therefore, the toy rocket is 19.1 meters above the ground when it explodes.