A 0.13 kg baseball moving at +25.80 m/s is slowed to a stop by a catcher who exerts a constant force of -362 N. How long does it take this...

Hello!

By Newton's Second law, the acceleration `a` of a baseball after contact with a catcher will be

`a=F/m,`

where `F` is the constant force and `m` is the mass of a ball. `a` is negative with respect to the direction of a ball. Constant acceleration (deceleration) make the speed `V` to decrease uniformly with the slope of `a:`

`V(t)=V_0-a*t,`

where `V_0` is the initial speed and `t` is the time since the beginning of a process.

Slowed to a stop means that the final speed is zero: `V(t_1)=0,` where `t_1` is the time in question. In such a way we obtain the equation for `t_1:`

`0=V_0-a*t_1,`

`t_1=V_0/a=V_0/((F/m))=(m*V_0)/F.`

In our case `m=0.13` kg, `V_0=25.80` `m/s` and `F=362` N.

The numerical result is `t_1=(0.13*25.80)/362 approx0.0093` (s).

Answer: it takes about 0.0093 s for a catcher to stop a ball.