Question

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 formula

For a ball thrown straight up, at the maximum height, the final velocity \( v_f = 0 \). We use the kinematic equation \( v_f^2=v_i^2 + 2a\Delta y \), where \( v_i \) is the initial velocity, \( a=-g\) (acceleration due to gravity, \( g = 9.8\ m/s^2\)) and \( \Delta y=h \) (the height we want to find).

Step2: Rearrange the formula to solve for \( h \)

From \( v_f^2=v_i^2+2(-g)h \), when \( v_f = 0 \), we get \( 0=v_i^2- 2gh \). Rearranging for \( h \), we have \( h=\frac{v_i^2}{2g} \).

Step3: Substitute the values

Given \( v_i = 10\ m/s \) and \( g = 9.8\ m/s^2 \), substitute into the formula: \( h=\frac{(10)^2}{2\times9.8}=\frac{100}{19.6}\approx5.1\ m \).

Answer:

The ball will go approximately \( 5.1\ m \) high.