Physics, 08.04.2020 02:12 kyleeb00

A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the right at 40.0 m/s; when it leaves the bat, the ball is traveling to the left at an angle of 30∘ above horizontal with a speed of 52.0 m/s. The ball and bat are in contact for 1.85 ms.

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A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the right at 40.0 m/s; when it leaves the bat, the ball is traveling to the left at an angle of 30∘ above horizontal with a speed of 52.0 m/s. The ball and bat are in contact for 1.85 ms, find the horizontal and vertical components of the average force on the ball.

The average horizontal force is 389.18 N and vertical component of the average force is 3221.62 N.

Explanation:

Given:

Mass of the baseball, = 0.145 kg

Velocity before the impact, v_1 = 40 m/s

Velocity after the impact, v_2 = 52 m/s

Angle of deflection, $\theta$ = 30 deg

Time of contact, $\triangle t$ = 1.85 ms

Conversion of milliseconds to seconds.

⇒ $\triangle t = 1.85\times 10^-^3\ sec$

Let the horizontal and vertical component of 'v1' and 'v2' be:

⇒ v_1x,v_1y,v_2x,v_2y

⇒ $v_1x=40\ ms^-^1$ and $v_2y= 0\ ms^-^1$

⇒ Taking the vector components of 'v2'

⇒ $v_2x=52cos(30) = 45.03\ ms^-^1$ and $v_2y=52sin(30)=40.97\ ms^-^1$

We have to find the impulse in x-y, direction.

Momentum with 'v1'

⇒ P_1_x=mv_1_x

⇒ P_1_x=0.145(40)

⇒ $P_1_x=5.8\ kg.m.s^-^1$ and P_1_y=0

Momentum with 'v2'

⇒ P_2_x=mv_2_x ⇒ P_2_y=mv_2_y

⇒ P_2_x=0.145(45.03) ⇒ P_2_y=0.145(40.97)

⇒ $P_2_x=6.52\ kg.m.s^-^1$ ⇒ $P_2_y=5.96\ kg.m.s^-^1$

Now the difference in terms of impulse (J).

Horizontally Vertically

⇒ J_x=P_2_x-P_1_x ⇒ J_y=P_2_y-P_1_y

⇒ J_x=6.52-5.8 ⇒ J_y=5.96-0

⇒ $J_x=0.72\ kg.m.s^-^1$ ⇒ $J_y=5.96\ kg.m.s^-^1$

Now we know that Impulse is the product of force and time.

So,

Horizontal force : Vertical force:

⇒ $(F_x)_a_v_g=\frac{J_x}{\triangle t}$ ⇒ $(F_y)_a_v_g=\frac{J_y}{\triangle t}$

⇒ $(F_x)_a_v_g=\frac{0.72\ kg.ms^-^1}{1.85\times 10^-^3\ s}$ ⇒ $(F_y)_a_v_g=\frac{5.96\ kg.m.s^-^1}{1.85\times 10^-^3\ s}$

⇒ $(F_x)_a_v_g=389.18\ N$ ⇒ $(F_x)_a_v_g=3221.62\ N$

So the horizontal force is 389.18 N and vertical component of the average force is 3221.62 N.

A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the rig

Answer from: Quest

low freq = 100 mhz

low wavelength =101.7m

length = 3x10^8/freq

Answer from: Quest

i had this question and the answer was 3.15 miles.

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