an object is thrown from a roof 200 feet above the ground, the height of the object in feet t seconds after being thrown is given by the function h(t) = 16 t^2 + 8t + 200

Question

What is the height of the object 2 seconds after being thrown
A 120 feet
B 152 feet
C 184 feet
D 192 feet

Answer 1

To find the height of the object 2 seconds after being thrown, we need to evaluate the function \( h(t) \) at \( t = 2 \).

Given the function: \[ h(t) = -16t^2 + 8t + 200 \]

we substitute \( t = 2 \): \[ h(2) = -16(2)^2 + 8(2) + 200 \]

Calculating step-by-step:

1. Calculate \( (2)^2 \): \[ (2)^2 = 4 \]

2. Multiply by -16: \[ -16 \times 4 = -64 \]

3. Calculate \( 8(2) \): \[ 8 \times 2 = 16 \]

4. Now substitute back into the equation: \[ h(2) = -64 + 16 + 200 \]

5. Simplify: \[ -64 + 16 = -48 \] \[ -48 + 200 = 152 \]

Therefore, the height of the object 2 seconds after being thrown is 152 feet, which corresponds to option B.