Question

question 1
1 pts
question 1: ball thrown straight up

1. if the initial velocity is 10 m/s, how high will the ball go?

Explanation:

Step1: Identify the physics concept

This problem involves a ball thrown straight up, so we can use the kinematic equation for vertical motion. At the maximum height, the final velocity \( v_f = 0 \, \text{m/s} \). The acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \) (acting downward, so we take it as negative in the upward direction). The initial velocity \( v_i = 10 \, \text{m/s} \). The kinematic equation we use is \( v_f^2 = v_i^2 + 2a\Delta y \), where \( \Delta y = h \) (the height we want to find), and \( a = -g = -9.8 \, \text{m/s}^2 \).

Step2: Rearrange the kinematic equation

We want to solve for \( h \) (which is \( \Delta y \)). Rearranging the equation \( v_f^2 = v_i^2 + 2a\Delta y \) for \( \Delta y \):

\[
\Delta y=\frac{v_f^2 - v_i^2}{2a}
\]

Substitute \( v_f = 0 \), \( v_i = 10 \, \text{m/s} \), and \( a=- 9.8 \, \text{m/s}^2 \):

\[
h=\frac{0-(10)^2}{2\times(- 9.8)}
\]

Step3: Calculate the height

First, calculate the numerator: \( 0 - 100=- 100 \)

Then, calculate the denominator: \( 2\times(-9.8)=-19.6 \)

Now, divide the numerator by the denominator:

\[
h=\frac{-100}{-19.6}\approx5.10 \, \text{m}
\]

Answer:

The ball will go approximately \( \boldsymbol{5.10 \, \text{meters}} \) high. (If we use \( g = 10 \, \text{m/s}^2 \) for approximation, the calculation would be \( h=\frac{0 - 10^2}{2\times(-10)}=\frac{-100}{-20} = 5 \, \text{m} \), but using \( g = 9.8 \, \text{m/s}^2 \) gives a more accurate result of approximately \( 5.10 \, \text{m} \).)