Question

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?

Answer 1

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)}`

`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  )/(g)\\h=v_(0)sin\alpha  .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`