Question

A toy rocket is launched from a platform 29 feet above the ground at a speed of 79 feet per second. The height of the rocket in feet is given by the polynomial −16t^2 + 79t + 29, where t is the time in seconds. How high will the rocket be after 3 seconds?

Answer 1

:)

Height = -16t² + 79t + 29

3 sec means t = 3, we substitute t with 3

H = -16(3)² + 79(3) + 29
    = -144 + 237 +29 
    = 122 feet

Answer 2

Answer:

$h(3)=122ft$

Step-by-step explanation:

The problem gives you everything, basically you only have to replace the given value, which is t=3 into the equation:

$h(t)=-16t^2+79t+29$

But, let's do a deeper analysis. This a projectile motion problem. For this kind of problems there are some fundamental equation. For example, we need to find the vertical displacement of the rocket. So the equation for the vertical displacement in projectile motion is given by:

$h_f-h_o=v_ot-\frac{1}{2}gt^2$  (1)

Where:

$h_f=Final\hspace{3}height\\h_o=Initial\hspace{3}height\\t=Time\\v_o=Initial\hspace{3}velocity\\g=Gravitational\hspace{3}acceleration$

Isolating  $h_f$ and rearranging the equation (1) we get:

$h_f=-\frac{1}{2}gt^2+v_ot+h_o$

Looks familiar? It has exactly the same form of the equation given by the problem. As you can see, if you replace the values of the speed and the initial position given by the problem, we got the same equation. Besides gravitational acceleration is a constant and its value is approximately 32ft/s^2, dividing it by 2, we get 16. Having this in mind, let's resolve the problem:

$h(3)=-16(3)^2+79(3)+29=-144+237+29=266-144=122ft$