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In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 19.0 m/s at an angle of 40.0° below the horizontal. The batter hits the ball toward center field, giving it a velocity of 50.0 m/s at 30.0° above the horizontal.

(a) Determine the impulse delivered to the ball.
N·s (magnitude)
How many degrees ° (above the horizontal)

(b) If the force on the ball increases linearly for 4.00 ms, holds constant for 20.0 ms, then decreases linearly to zero in another 4.00 ms, what is the maximum force on the ball?
N (magnitude)
How many degrees? (above the horizontal)

Respuesta :

Answer:

Explanation:

We shall write the initial and final velocity after impulse in vector form

initial velocity

V₁ = 19 cos 40 i - 19 sin 40 j

V₁ = 14.55 i - 12.21  j

Similarly velocity after the impulse

V₂ = 50 cos 30 i + 50 sin30 j

= 43.30 i + 25 j

Now V₂-V₁ , change in velocity

= 43.30 i + 25 j - 14.55 i + 12.21  j

= 28.75 i +37.21 j

change in momentum

maas x change in velocity

.2 (28.75 i +37.21 j )

5.75 i + 7.442 j

Impulse = change in momentum

= 5.75 i + 7.442 j

magnitude

= √5.75² +7.442²

= 9.4 kg m/s

TanФ = 7.442 / 5.75

Ф = 52 °

b )

total time = 28 x 10⁻³ s

Force x time = change in momentum

force = change in momentum / time

= 9.4 / 28 x 10⁻³

= 335.7 N.

Direction of force will be the same as before ie 52 degree from the horizontal .

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