Question

A ball is thrown upward at a velocity of 19.6 m/s. What is its velocity after 3.0, assuming negligible air resistance?

Answer 1

If the ball travels 19.6 meters every second and it travels for 3 seconds, obviously you have to multiply the velocity by 3:

19.6 x 3= 58.8m/s

Answer 2

Answer:

After 3 seconds the velocity becomes 49 m/s

Explanation:

Applying upward displacement equation:

$Y = V_yt +\frac{1}{2}gt^2$

differentiate this equation,

$\frac{dy}{dt} = V_y + gt$

where; dy/dt is the velocity function;

After 3 seconds, the velocity becomes;

V = 19.6 + (9.8)(3)

V = 19.6 + 29.4

V = 49 m/s

Therefore, after 3 seconds the velocity becomes 49 m/s