A major leaguer hits a baseball so that it leaves the bat at a speed of 30.5m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 30.5m/s and at an angle of36.1? above the horizontal. You can ignore air resistance.At what two times is the baseball at a height of 10.0m above the point at which it left the bat?Calculate the horizontal component of the baseball's velocity at each of the two times you found in part A. Calculate the vertical component of the baseball's velocity at each of the two times you found in part A. What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat? What is the direction of the baseball's velocity when it returns to the level at which it left the bat?

Explanation / Answer

i) from equation of motion in vertical(y) direction:     ( here uy = component of velocity in y-direction)

Sy = uy*t + (1/2)*ay*t2                            ( Sy = displacement in y- direction, t = time,

                                                             ay = acceleration in Y-direction= - 9.8 m/s2, uy = u*sin(36.1) )

10= u*sin(36.1)*t - (1/2)*g*t2

10 = 30.5*sin(36.1)*t - (1/2)*9.8*t2

on solving this equation

t = 2.98 sec,   0.68 sec

(ii) horizontal component of velocity at any time:

Vx = ux +a*t

     = 30.5*cos(36.1) + 0                           ( a=0 since velocity in horizontal direction will remain constant)

     = 24.64 m/s                  ( for both the times)

(iii) vertical component of baseball velocity :

Vy = uy + ay*t

Vy    = u*sin(36.1) - g*t                           ---------------------1

at t= 2.98 sec, from eq 1

Vy = -11.23 m/s

at t= 0.68 sec, from eq 1

Vy = 11.306 m/s

(iv) Because their is no air resistance ,the flight will be parabolic and symmteric about mid point.

So speed when it is at the same levelwhen it left the bat = 30.5 m/s

(v) direction , at an angle of 36.10 below horizontal

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