To find the maximum height of the toy rocket, we need to find the vertex of the quadratic equation h(t) = -4.9t² + 19.6t + 0, where a = -4.9 (half of the gravitational constant since the rocket is launched upwards) and v = 19.6.
The formula for the vertex of a quadratic equation ax² + bx + c is given by t = -b/2a.
In this case, t = -19.6 / 2(-4.9) = 2 seconds.
Now, plug the value of t back into the equation h(t) to find the maximum height:
h(2) = -4.9(2)² + 19.6(2) + 0
h(2) = -4.9(4) + 39.2
h(2) = -19.6 + 39.2
h(2) = 19.6 meters
Therefore, the toy rocket is 19.6 meters above the ground when it explodes.
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) = 1/2at² + vt + 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 height in meters modeled as a function of time, t.
Due to a malfunction, the toy rocket explodes when it reaches its maximum height. How high abov the ground is the toy rocket when it explodes? Round to the nearest tenth.
3 answers
well all you really need to solve is the vertical problem
h(t) = 1/2at² + vt + s
v(t) = Vinitialup - g t = 0 at top
0 = 19.6 - 9.8 t
t = 2 seconds flying upward
then
h = (1/2) (-9.8) *4 + 19.6 * 2 + 0.5 *note it started 1/2 meter up
= -19.6 +39.2 + .5
= 20.1
h(t) = 1/2at² + vt + s
v(t) = Vinitialup - g t = 0 at top
0 = 19.6 - 9.8 t
t = 2 seconds flying upward
then
h = (1/2) (-9.8) *4 + 19.6 * 2 + 0.5 *note it started 1/2 meter up
= -19.6 +39.2 + .5
= 20.1
I apologize for the confusion in my previous response.
The toy rocket is 20.1 meters above the ground when it explodes.
The toy rocket is 20.1 meters above the ground when it explodes.