8. A rocket is launched vertically upward from the ground. Its engine accelerates the rocket for 10 s from launch until it reaches a velocity of \( 250 \mathrm{~m} \cdot \mathrm{~s}^{-1} \). After 10 s turns its engine off. a) What is the acceleration of the rocket during the first 10 s ? b) What is the acceleration after 10 s ? c) Calculate the maximum height reached by the rocket.

To solve the problem, we will break it down into parts based on the questions asked. ### Given Information: - Initial velocity (\( u \)) = 0 m/s (the rocket starts from rest) - Final velocity after 10 seconds (\( v \)) = 250 m/s - Time of acceleration (\( t \)) = 10 s ### a) What is the acceleration of the rocket during the first 10 s? We can use the formula for acceleration (\( a \)): \[ a = \frac{v - u}{t} \] Substituting the known values: \[ a = \frac{250 \, \text{m/s} - 0 \, \text{m/s}}{10 \, \text{s}} = \frac{250 \, \text{m/s}}{10 \, \text{s}} = 25 \, \text{m/s}^2 \] ### b) What is the acceleration after 10 s? After the engine turns off, the only force acting on the rocket is gravity, which will decelerate the rocket. The acceleration due to gravity (\( g \)) is approximately \( 9.81 \, \text{m/s}^2 \) downward. Therefore, the acceleration after 10 seconds is: \[ a = -g = -9.81 \, \text{m/s}^2 \] ### c) Calculate the maximum height reached by the rocket. To find the maximum height, we need to calculate the height reached during the first 10 seconds of acceleration and then the height gained after the engine turns off until the rocket comes to a stop. 1. **Height during the first 10 seconds**: We can use the formula for displacement (\( s \)): \[ s = ut + \frac{1}{2} a t^2 \] Substituting the known values: \[ s_1 = 0 \cdot 10 + \frac{1}{2} \cdot 25 \cdot (10)^2 = \frac{1}{2} \cdot 25 \cdot 100 = 1250 \, \text{m} \] 2. **Height after the engine turns off**: The rocket will continue to rise until its velocity becomes 0. We can use the following kinematic equation: \[ v^2 = u^2 + 2as \] Here, \( v = 0 \, \text{m/s} \) (at the maximum height), \( u = 250 \, \text{m/s} \) (initial velocity after 10 seconds), and \( a = -9.81 \, \text{m/s}^2 \). Rearranging the equation to solve for \( s \): \[ 0 = (250)^2 + 2(-9.81)s \] \[ 0 = 62500 - 19.62s \] \[ 19.62s = 62500 \] \[ s_2 = \frac{62500}{19.62} \approx 3185.1 \, \text{m} \] 3. **Total maximum height**: The total maximum height (\( H \)) is the sum of the heights from both phases: \[ H = s_1 + s_2 = 1250 \, \text{m} + 3185.1 \, \text{m} \approx 4435.1 \, \text{m} \] ### Final Answers: a) The acceleration of the rocket during the first 10 s is \( 25 \, \text{m/s}^2 \). b) The acceleration after 10 s is \( -9.81 \, \text{m/s}^2 \). c) The maximum height reached by the rocket is approximately \( 4435.1 \, \text{m} \).