We appreciate your visit to A toy rocket is launched straight upward from the ground with an acceleration of tex 20 0 text m s 2 tex After tex 3. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Final answer:
The maximum height achieved by the toy rocket is 312.1 meters by considering two stages of the rocket's flight: while the rocket's fuel was burning and accelerating the rocket upwards, and after the fuel was exhausted, during which the rocket decelerated under gravity to achieve maximum altitude before falling back.
Explanation:
In order to calculate the maximum height achieved by a rocket, we need to consider two stages separately: first, while the rocket's fuel is burning and accelerating the rocket upward, and second, when the rocket is decelerating upward and finally falling back to the ground under the influence of gravity alone.
Stage 1: Rocket's fuel burning (0s to 3.20s)
The distance travelled by the rocket while its fuel was burning can be calculated using the formula for the distance under constant acceleration, which is d = ut + 0.5at^2. Here, 'u' is the initial velocity, 'a' is the acceleration, 't' is the time. Since the rocket started from rest, u=0. Hence we have d1 = 0.5*a*t^2 = 0.5 * 20 * (3.20^2) = 102.4 meters.
Stage 2: Rocket's fuel finishes burning (3.20s to max height)
When the fuel is finished, the rocket still carries the velocity it had at the end of 3.20 seconds, which can be calculated as v = u + at = 0 + (20*3.20) = 64 m/s. This velocity will carry the rocket upwards further after the fuel has been exhausted till it decelerates to zero velocity under gravity. The maximum height (d2) achieved during this stage can be calculated using the third equation of motion v^2 = u^2 + 2as, where 'u' is initial velocity, 'v' is final velocity, 'a' is acceleration, and 's' is displacement / distance. Rearranging it gives s = (v^2 - u^2) / (2a). Here, u = 64 m/s, v = 0 (at max height, final velocity is zero), and a = -9.8 m/s^2 (acceleration due to gravity is downwards). Substituting these values gives d2 = (0 - (64^2)) / ( 2*(-9.8)) = 209.7 meters.
The maximum height achieved is the sum of the distances travelled in both stages, which is d1 + d2 = 102.4 + 209.7 = 312.1 meters.
Learn more about Rocket maximum height here:
https://brainly.com/question/32241391
#SPJ4
Thanks for taking the time to read A toy rocket is launched straight upward from the ground with an acceleration of tex 20 0 text m s 2 tex After tex 3. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
