A toy rocket is shot upward into the air from a height of 1/2 meter above the ground with an initial of velocity of 19.6 meters per second. The formula for the vertical motion of an object is h(t) = 1/2at^2+vt+s, where the gravitational constant, a, is -9.8 m/s squared, is the initial velocity, S is the initial height, and h(t) is the height in meters models as a function in 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 of the toy rocket when it explodes, we need to first find the time it takes for the rocket to reach its maximum height.
At the maximum height, the velocity of the rocket will be 0. So we can use the formula v = at + v where a = -9.8 m/s^2 and v = 19.6 m/s to solve for the time it takes to reach the maximum height.
0 = -9.8t + 19.6
9.8t = 19.6
t = 2 seconds
Now that we have the time it takes to reach the maximum height, we can use the equation h(t) = 1/2at^2 + vt + s to find the height at t = 2 seconds.
h(2) = 1/2(-9.8)(2)^2 + (19.6)(2) + 0.5
h(2) = -19.6 + 39.2 + 1
h(2) = 20.6 meters
Therefore, the toy rocket is 20.6 meters above the ground when it explodes.