Question
A ball is thrown directly upward from the ground with an initial velocity of 45.1ft./sec . Represent the height of the ball from the ground t seconds after it was thrown upward using a model h(t)=−16(t)^2+v0t+s0 , then use the model to find the height of the ball 2 seconds after it was thrown.
Answers
GPT-4o mini
To model the height \( h(t) \) of the ball thrown directly upward, we can use the equation:
\[
h(t) = -16t^2 + v_0t + s_0
\]
Where:
- \( v_0 \) is the initial velocity,
- \( s_0 \) is the initial height (which is 0 since it is thrown from the ground).
Given:
- \( v_0 = 45.1 \, \text{ft/sec} \)
- \( s_0 = 0 \, \text{ft} \)
Substituting these values into the equation, we have:
\[
h(t) = -16t^2 + 45.1t + 0
\]
This simplifies to:
\[
h(t) = -16t^2 + 45.1t
\]
Next, we will calculate the height of the ball 2 seconds after it was thrown. We will substitute \( t = 2 \) into the equation:
\[
h(2) = -16(2^2) + 45.1(2)
\]
\[
= -16(4) + 90.2
\]
\[
= -64 + 90.2
\]
\[
= 26.2 \, \text{ft}
\]
Thus, the height of the ball 2 seconds after it was thrown is **26.2 feet**.
\[
h(t) = -16t^2 + v_0t + s_0
\]
Where:
- \( v_0 \) is the initial velocity,
- \( s_0 \) is the initial height (which is 0 since it is thrown from the ground).
Given:
- \( v_0 = 45.1 \, \text{ft/sec} \)
- \( s_0 = 0 \, \text{ft} \)
Substituting these values into the equation, we have:
\[
h(t) = -16t^2 + 45.1t + 0
\]
This simplifies to:
\[
h(t) = -16t^2 + 45.1t
\]
Next, we will calculate the height of the ball 2 seconds after it was thrown. We will substitute \( t = 2 \) into the equation:
\[
h(2) = -16(2^2) + 45.1(2)
\]
\[
= -16(4) + 90.2
\]
\[
= -64 + 90.2
\]
\[
= 26.2 \, \text{ft}
\]
Thus, the height of the ball 2 seconds after it was thrown is **26.2 feet**.