Question

question 17 of 27
how fast is a ball going when it hits the ground after being dropped from a height of 9 m? the acceleration of gravity is 9.8 m/s².

a. 13.3 m/s
b. 17.6 m/s
c. 11.5 m/s
d. 14.9 m/s

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v^{2}=v_{0}^{2}+2ax$. The ball is dropped, so the initial velocity $v_{0} = 0\ m/s$, the acceleration $a = g=9.8\ m/s^{2}$ and the displacement $x = h = 9\ m$.

Step2: Substitute the values into the equation

Substitute $v_{0}=0\ m/s$, $a = 9.8\ m/s^{2}$ and $x = 9\ m$ into $v^{2}=v_{0}^{2}+2ax$. We get $v^{2}=0 + 2\times9.8\times9$.

Step3: Calculate the value of $v^{2}$

$v^{2}=2\times9.8\times9=176.4$.

Step4: Solve for $v$

$v=\sqrt{176.4}\approx 13.3\ m/s$.

Answer:

A. 13.3 m/s