Question

A baseball pitcher throws a fastball which leaves the pitcher’s hand traveling with an initial speed of 110 ft/sec(approximately 75 mph), and an initial height of 6 feet. Ignoring wind resistance, the ball’s trajectory (path) is described by the equations: x=(110cos(theta))t y=6 + (110sin(theta))t - 16t^2

Answer 1

A.) If the pitcher throws the ball directly on the horizontal, how long does it take for the ball to cross the home plate,
60.5 feet from the pitchers mound? (give your answer correct to one decimal place)

Answer 2

B.) If the pitcher throws the ball directly on the horizontal, at what height does the ball cross the home plate? (give
your answer correct to one decimal place

Answer 3

A) Let x= 60.5 and solve for t.

B) Use the value of t you got in A) above to find the value of y.

Note that if the ball is released exactly level, cos (theta) =1 and sin (theta)=0.

Answer 4

thank you for your reply.
so for A.) i got .55 sec and B.) 1.16 ft? is that right?

Answer 5

one more question.
If the ball crosses home plate between 1.5 and 4.5 feet above the ground it is considered a “strike”. What is the range of theta , correct to the nearest 0.01 of a degree, for which the pitch will be considered a strike?

Answer 6

I think your answers are correct. To find the maximum and minimum angles of theta, put 1.5 in for y, and work back to get theta. Then, repeat the process with y=4.5. At the small angle differences involved, the time shouldn't change enough to make any difference.