A 0.140 kg baseball traveling 35.0 m/s strikes the catchers mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What was the average force applied by the ball to the glove?

The average force corresponds to constant deceleration. With constant deceleration the average speed in the mitt is 35/2 = 17.5m/s. Therefore the time taken to cover the 11 cm (=0.11m ) = T = 0.11/17.5 = 0.006286 secs.

Average acceleration a = v/t = 35/0.006286 = 5568 m/s²

So average force = m.a = 0.14 x 5568 = 780 N – that’s reason enough to wear a glove!

First, determine the deceleration of the ball as it comes to rest after hitting the mitt:

v^2 = u^2 + 2as,

where v = 0, u =35, s = 0.11

After you have calculated a, you can get the average force from

f = ma

Impulse I = m*V = F*t

he know

m = 0.14kg

V = 35.0 m/sec

their product equates :

I = 0.14*35.0 = 4.90 kgm/sec

…we know, furthermore , the space S to rest given by this formula :

S = 1/2*V*t

solving by t :

t = 2*S/V = (2*11/100)/35.0 = 0.00629 sec (or 6.29*10^-3)

and finally :

F = I/t = 4.90*10^3/6.29 = 780 N (80 kg approx)

I’m sorry I don’t know about this

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