Consider two identical baseballs which start at the same location; their terminal speed at this location is 32 m/s. At t = 0s, Ball A is thrown straight up at m/s; Ball B is thrown at 16m/s 30.0° below the horizontal at 32 m/s. How do the magnitudes of the baseballs' accelerations32 2at t = 0s compare (do not neglect air resistance)?

Complete Question

The complete question is shown on the first uploaded image

The correct option is  B

Explanation:

From the question we are told that

  The terminal  velocity of both balls is  

   The speed with which ball A is thrown up at t =0 s is  

   The speed with which ball B is thrown 30° below the horizontal  at t =0 s is  

Generally the acceleration of ball A is (g i.e acceleration due gravity)given that it is thrown straight up  

Generally the vertical  acceleration of ball B is mathematically represented as

Generally the ratio of both acceleration is  

=>

=>  

=>

Rationalizing

=>

=>

=>

im guessing the west of station b

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s