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Lauren launches a toy rocket from a platform. The height of the rocket in feet is given by 
h(t)=-16t^(2)+48 t+28 where 
t represents the time in seconds after launch. What is the appropriate domain for this situation?
Answer:

Lauren launches a toy rocket from a platform. The height of the rocket in feet is given by h(t)=16t2+48t+28 h(t)=-16 t^{2}+48 t+28 where t t represents the time in seconds after launch. What is the appropriate domain for this situation?\newlineAnswer:

Full solution

Q. Lauren launches a toy rocket from a platform. The height of the rocket in feet is given by h(t)=16t2+48t+28 h(t)=-16 t^{2}+48 t+28 where t t represents the time in seconds after launch. What is the appropriate domain for this situation?\newlineAnswer:
  1. Define Domain: The domain of a function is the set of all possible input values (in this case, time tt) for which the function is defined. Since the function h(t)h(t) represents the height of a rocket over time, the domain will be restricted to the time interval from the launch of the rocket until it hits the ground.
  2. Find Rocket's Impact Time: To find when the rocket hits the ground, we need to determine when the height h(t)h(t) is equal to zero. This involves solving the quadratic equation 16t2+48t+28=0-16t^2 + 48t + 28 = 0 for tt.
  3. Apply Quadratic Formula: We can use the quadratic formula to solve for tt. The quadratic formula is t=b±b24ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=16a = -16, b=48b = 48, and c=28c = 28.
  4. Calculate Discriminant: First, we calculate the discriminant, which is b24acb^2 - 4ac. Plugging in the values, we get 4824(16)(28)48^2 - 4(-16)(28).
  5. Simplify Discriminant: Calculating the discriminant gives us 2304(1792)2304 - (-1792), which simplifies to 2304+1792=40962304 + 1792 = 4096.
  6. Apply Quadratic Formula: Now we can apply the quadratic formula. The two possible solutions for tt are t=48±409632t = \frac{-48 \pm \sqrt{4096}}{-32}.
  7. Calculate Possible Times: The square root of 40964096 is 6464. So the two possible solutions for tt are t=(48±64)/(32)t = (-48 \pm 64) / (-32).
  8. Discard Negative Time: This gives us two possible times: t=48+6432t = \frac{-48 + 64}{-32} and t=486432t = \frac{-48 - 64}{-32}.
  9. Determine Impact Time: Calculating these, we get t=16(32)t = \frac{16}{(-32)} and t=112(32)t = \frac{-112}{(-32)}, which simplify to t=0.5t = -0.5 and t=3.5t = 3.5, respectively.
  10. Establish Domain: Since time cannot be negative in this context, we discard the negative value. Therefore, the rocket will hit the ground after 3.53.5 seconds.
  11. Establish Domain: Since time cannot be negative in this context, we discard the negative value. Therefore, the rocket will hit the ground after 3.53.5 seconds.The appropriate domain for the function h(t)h(t) is from the time of launch, t=0t = 0, until the rocket hits the ground, t=3.5t = 3.5 seconds. Therefore, the domain is [0,3.5][0, 3.5].

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