Can anyone help me here? I have a test tomorrow similar question to this?

Question

A baseball thrown in the air at an angle of 0.5% below the horizontal at 90 miles per hour towards home plate which is 60 feet from the pitching rubber. the pitcher releases the ball 5 feet above the ground and 4 feet in front of the pitching rubber. recall: 1 mile=5280 ft

a) write the parametric equations for the path of the ball

b) how many seconds it takes to cross home plate

c)Give the height of the pitch when it crosses home plate

Answer 1

Start by computing the horizontal component of velocity:

The angle is .5/100
a)
The horizontal distance is
cos(.5/100)*90*5280/3600 ft/sec *t

The vertical distance is
(-sin(.5/100)*90*5280/3600-1/2*32*t)ft/sec*t

b) the horizontal distance to home plate from release point is 56 ft
t=56/(cos(.5/100)*90*5280/3600) sec
=.424 secs

Knowing the time to the plate, plug into the vertical distance subtracted from 4 ft
h=4-(sin(.5/100)*90*5280/3600+.5*32*.424)*.424
=1.12 ft

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