Question

A boy throws a ball with an initial velocity of 19.6 m/s. What maximum height does the ball reach?

Vf^2=vi^2+2gs

Answer 1

Hello!

The answer is: 19.59 m

Why?

Since there is no information about the launch type, we can assume that the ball is thrown vertically upward.

When the ball reaches the maximum height, just at that moment, the velocity turns to 0, and after that moment, the ball starts falling, so:

We will use the following formula:

$Vf^2=Vi^2+2*g*s$

Where:

Vf= Final velocity = 0

Vi= Initial velocity = $\frac{19.6m}{s}$

g = Gravity Acceleration = $\frac{9.81m}{s^{2} }$

s = Traveled distance

$0=19.6^2+2*-9.81*s\\s=\frac{19.6^2}{2*9.81}=\frac{384.16}{19.62}=19.59m$

Have a nice day!