Question

A baseball is hit when it is 2.5 ft above the ground. It leaves the bat with an initial velocity of 145 ft/sec at a launch angle of 23°. At the instant the ball is hit, an instantaneous gust of wind blows against the ball, adding a component of -14i (ft/sec) to the ball’s initial velocity. A 15-ft-high fence lies 300 ft from home plate in the direction of the flight.
a. Find a vector equation for the path of the baseball.
b. How high does the baseball go, and when does it reach maxi-mum height?
c. Find the range and flight time of the baseball, assuming that the ball is not caught.
d. When is the baseball 20 ft high? How far (ground distance) is the baseball from home plate at that height?
e. Has the batter hit a home run? Explain.

Answer 1

Answer:

Explanation:

Take base of the ground as origin .

component of initial velocity along i and j direction is 145 con23 and 145 sin23 . Along j , gravity acts but along i , no force acts .  

The path  of ball in vector form

s = (145 cos23- 14 )t  i + ( 2.5 + 145sin23 t - 1/2 g t² ) j

t is time period .

b )

vertical component of initial velocity = 145 sin 23 =

for vertical displacement

v² = u² - 2gH

For maximum height , v = 0

0 = (145 sin 23 )² - 2 g H , H is maximum height attained .

H = 3209.56 / 2 x 9.8

= 163.75 m

Total height attained = 163.75 + 2.5 = 166.25 m

if time be t for reaching maximum height

v = u -gt

0 = 145 sin 23 - gt

t = 145 sin23 / g

= 5.78 s

c )

For time of flight , vertical displacement = 2.5 m

2.5 = - 145 sin 23 t + 1/2 g t²

2.5 = -56.65 t + 4.9 t²

4.9 t² - 56.65 t - 2.5 = 0

t = 11.60s

horizontal displacement during this period = 145 cos23 x 11.60 = 1548.28 m

Range = 1548.28 m.