Question

the gravitational force exerted on a baseball is 2.21 N down, a pitcher throws the ball horizontally with velocity 18.0 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 170 ms. the ball starts from rest. Through what distance does it move before its release? what are the magnitude and direction of the force of the pitcher exerts on the ball?

Answer 1

To have a weight of 2.21N., the ball's mass is (2.21/9.8) = .226kg. 
a) d = 1/2 (vt), = 1/2 (18 x .17), = 1.53m. 
b) Acceleration of the ball = (v/t), = 18/.17, = 105.88m/sec^2. 
f = (ma), = .226 x 105.88, = 23.92N. 

Answer 2

Hello!

The gravitational force exerted on a baseball is 2.21 N down, a pitcher throws the ball horizontally with velocity 18.0 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 170 ms. the ball starts from rest. Through what distance does it move before its release? what are the magnitude and direction of the force of the pitcher exerts on the ball ?

We have the following data:

P (force-weight or gravitacional force) = 2.21 N

m (mass) = ? (in kg)

g (acceleration of gravity) ≈ 9.8 m/s²

Vf (final velocity) = 18 m/s

Vi (initial velocity) = 0 m/s (in rest)

Δt (time interval) = 170 ms = 0.17 s

a (acceleration) = ? (in m/s²)

F (force or magnitude) = ? (in N or kg.m/s²)

[First Step]  Let's find the mass of the ball, let's see:

$P = m*g$

$2.21 = m * 9.8$

$2.21 = 9.8\:m$

$9.8\:m = 2.21$

$m = \dfrac{2.21}{9.8}$

$\boxed{m \approx 0.226\:kg}$

[Second Step] Let's find the acceleration, let'see:

$a = \dfrac{\Delta{V}}{\Delta{T}}$

$a = \dfrac{V_f - V_i}{\Delta{T}}$

$a = \dfrac{18.0\:m/s - 0\:m/s}{0.17\:s}$

$a = \dfrac{18\:m/s}{0.17\:s}$

$\boxed{a \approx 105.88\:m/s^2}$

[Third Step] Through what distance does it move before its release?

By the formula of the space of the Uniformly Varied Movement, it is:

$d = V_i*t + \dfrac{a*t^2}{2}$

$d = 0*0.17 + \dfrac{105.88*0.17^2}{2}$

$d = 0 + \dfrac{105.88*0.0289}{2}$

$d = \dfrac{3.059932}{2}$

$\boxed{\boxed{d \approx 1.53\:m}}\:\:\:\:\:\:\bf\blue{\checkmark}$

*** [Fourth Step] What are the magnitude....

$F =m*a$

$F = 0.226\:kg*105.88\:m/s^2$

$F = 23.92888\:kg*m/s^2$

$\boxed{\boxed{F \approx 24\:N}}\:\:\:\:\:\:\bf\green{\checkmark}$

***.... and direction of the force of the pitcher exerts on the ball ?

$tan\:C= \dfrac{F_1\downarrow}{F_2\rightarrow}$

$tan\:C= \dfrac{2.21}{24}$

$tan\:C \approx 0.093 \to \boxed{\boxed{C = arctan\:0.093 \approx 5.3\:\º}}\Longrightarrow(above\:the\:horizontal)\:\:\:\:\:\:\bf\green{\checkmark}$

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