33.8k views
5 votes
Use the model for projectile motion, assuming no air resistance.

A baseball, hit above the ground, leaves the bat at an angle of 45 degrees and is caught by an outfielder 3 feet above the ground and 300 feet from home plate.

What is the initial speed of the ball, and how high does it rise?

1 Answer

6 votes

Answer:

s=29.93m/s

h=22.88m

Explanation:

we must find the initial speed, we will determine its position (x-y).

x component
s=0+v_(0)cos\alpha.t+0=v_(0)cos\alpha.t

y component
h=0+v_(0)sin\alpha.t-(1)/(2)gt^(2)=v_(0)sin\alpha.t-(1)/(2)gt^(2)\\ since the ball is caught at the same height then h=0


h=v_(0)sin\alpha.t-(1)/(2)gt^(2)=0\\v_(0)sin\alpha.t-(1)/(2)gt^(2)=0;v_(0)sin\alpha.t=(1)/(2)gt^(2)\\t=(2v_(0)sin\alpha)/(g)\\

where t= flight time;
s=v_(0)cos\alpha.t, replacing t:


v_(0)=\sqrt{(sg)/(sin2\alpha)}


s=v_(0)cos\alpha((2sin\alpha )/(g))=(v_(0) ^(2)2sin\alpha.cos\alpha)/(g)=(v_(0) ^(2)sin(2\alpha))/(g)]</p><p>: the values ​​must be taken to the same units</p><p></p><p>[tex]300ft*0.3048m/ft=91.44m


v_(0)=\sqrt{(91.44m*9.8(m)/(s^(2)))/(sin2(45))}=\sqrt{895.112((m)/(s) )^(2) }=29.93(m)/(s)

To calculate the height you should know that this is achieved when its component at y = 0


v_(y)=v_(0)sin\alpha-gt=0;gt=v_(0)sin\alpha\\\\ t=(v_(0)sin\alpha &nbsp;)/(g)\\h=v_(0)sin\alpha &nbsp;.t-(1)/(2)gt^(2)

replacing t;
h=v_(0)sin\alpha((v_(0)sin\alpha)/(g))-(1)/(2)g((v_(0sin\alpha))/(g)) ^(2)\\

finally


h=((v_(0)sin\alpha)^(2))/(2g)=((29.95*sin45)^(2))/(2*9.8)=22.88m

User Dmitry Lomov
by
6.9k points