Question

A ball leaves a players foot at an initial velocity of 2.12m/s horizontally towards the goal and slows down at a rate of -0.82 m/s/s. How fast is the ball going after 0.15 seconds.

Answer 1

@Hoslos

Answer 2

According to motion formulas, the appropriate one will be a=(v-u)/t, where a is the acceleration, v is the final speed, u is the initial speed and t is the time taken. Now we simply replace values and make v subject of the formula or vice-versa. So:
v= u-at = 2.12-(0.82*0.15)= 1.997m/s.