Question

a person throws a ball up into the air at an initial speed of 22 m/s. what is the balls velocity 4.1 seconds after it is thrown

Answer 1

a person throws a ball up into the air at an initial speed of $22\mathrm{m / s}$ . what is the balls velocity 4.1 seconds after it is thrown

Solution:

$\mathrm{V}_{1} = 22\frac{\mathrm{m}}{\mathrm{s}} -$ initial velocity

$\mathrm{V}_{2} -$ velocity 4.1 seconds after it is thrown

The rate equation for the ball:

$$y: V _ {2} = V _ {1} - g t = 2 2 \frac {m}{s} - 9. 8 \frac {m}{s ^ {2}} \cdot 4. 1 s = - 1 8. 8 \frac {m}{s}$$

A minus sign shows that the speed has changed its direction, and now it is directed vertically downward.

Answer: the balls velocity 4.1 seconds after it is thrown is $18.8\frac{\mathrm{m}}{\mathrm{s}}$ and it is directed vertically downward (opposite to initial velocity).