A player hits a baseball, which has an initial velocity of 42 m/s. At the moment

Question

A player hits a baseball, which has an initial velocity of 42 m/s. At the moment the ball is hit with the bat, the height of the baseball will be 0.22 m. The distance from the point of contact to the wall is 170 m. The baseball has an angle of 40 degrees with respect to the horizontal. the player will get a homeun, if he or she hits the baseball pass the wall. The height of the wall is 6m.

a) will the player get a homerun? if so, the by how much does the baseball clear the wall?

b) Now assume that the wall is completely removed. Find the horizontal range and maximum height of the baseball. Also assume that the path of the baseball is symmetical, so it lands at a location that is 0.22 m above the ground.

Explanation / Answer

a) Start off by making a parametric equation x=Vcos()t and y=yi+Vsin()t-1/2gt2

inserting in variables into the first equation 170m=42m/s cos(40)t and solving for t t=5.283791409...s.

Plug this into the second equation along with variables and y=.22m+42m/s sin(40)5.283791409s-4.9m/s2(5.283791409)=6.066524186m since this is greater than 6 it's a homerun and he does it by 0.0665241859m.

b) Now for this one you only change the y= equation so there is no yi value. Now set that equation equal to zero because there is no net displacement and solve for t.

0=42m/s sin(40)t-4.9m/s2t2 and you get t to equal 5.509608083s. Now plug this t into the first equation x=42m/s cos(40)5.509608083s=177.2653955.

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