Question

Solve the problem. Round to the nearest​ tenth, if necessary.
If an object is propelled upward from ground level with an initial velocity of 50.3 feet per​ second, its height h in feet t seconds later is given by the equation h equals negative 16 t squared plus 50.3 t. After how many seconds does the object hit the​ ground?
Question content area bottom
Part 1
A.1.6
1.6
B.6.3
6.3
C.
.3
D.3.1
3.1

Answer 1

To find out when the object hits the ground, we need to determine when the height \( h \) equals zero. The height \( h \) as a function of time \( t \) is given by the equation:

\[
h(t) = -16t^2 + 50.3t
\]

We set \( h(t) = 0 \) to find the time when the object hits the ground:

\[
-16t^2 + 50.3t = 0
\]

This equation can be factored:

\[
t(-16t + 50.3) = 0
\]

This gives us two solutions:

1. \( t = 0 \) (the time when the object is launched, which we already know)
2. \( -16t + 50.3 = 0 \)

Now, solving for \( t \):

\[
-16t + 50.3 = 0 \\
16t = 50.3 \\
t = \frac{50.3}{16} \\
t \approx 3.14375
\]

Rounding this value to the nearest tenth, we get:

\[
t \approx 3.1 \text{ seconds}
\]

Thus, the object hits the ground after approximately **3.1 seconds**. Therefore, the correct answer is:

**D. 3.1**