Question

a ball is rolling down a hill at +1.0m/s the ball is uniformly accelerating at +4.9m/s^2 what will the balls velocity be 4.0 later

Answer 1

we are given

initial velocity is +1.0 m/s

so, $u=1.0m/s$

Let's assume final velocity as v

uniformly accelerating at +4.9m/s^2

so, $a=4.9m/s^2$

time is 4 sec

so, t=4 s

now, we can use formula

$a=\frac{v-u}{t}$

we can plug values

$4.9=\frac{v-1}{4}$

$4.9*4=4*\frac{v-1}{4}$

$4.9*4=v-1$

$19.6=v-1$

$v=20.6m/s$...............Answer

Answer 2

Answer:

So velocity of ball after 4 sec will be 20.6 m/sec

Step-by-step explanation:

We have given initial velocity of the ball u = 1 m/sec

Acceleration of the ball $a=4.9m/sec^2$

We have to find velocity after 4 sec

So ti,e t = 4 sec

According to first equation of motion we know that v = u+at

So $v=1+4.9\times 4=20.6m/sec$

So velocity of ball after 4 sec will be 20.6 m/sec