(a) The pitcher must throw the ball at 27.7 m/s
The momentum of an object is given by:
![p=mv](https://img.qammunity.org/2020/formulas/physics/middle-school/lldmrpmkc3i68kifbfb2c4hi3vblitcsch.png)
where
m is the mass of the object
v is the object's velocity
Let's calculate the momentum of the bullet, which has a mass of
m = 2.70 g = 0.0027 kg
and a velocity of
![v=1.50\cdot 10^3 m/s](https://img.qammunity.org/2020/formulas/physics/college/k48ntu0chkwjvynckji6c6yrqc0txvi7us.png)
Its momentum is:
![p=mv=(0.0027 kg)(1.50\cdot 10^(3) m/s)=4.05 kg m/s](https://img.qammunity.org/2020/formulas/physics/college/cfysca310ke8am718r01syjj5ydcsz4vgt.png)
The pitcher must throw the baseball with this same momentum. The mass of the ball is
m = 0.146 kg
So the velocity of the ball must be
![v=(p)/(m)=(4.05 kg m/s)/(0.146 kg)=27.7 m/s](https://img.qammunity.org/2020/formulas/physics/college/bk7ztoy8u14kh18qe89jpdue9loqbkr5m1.png)
So, the pitcher must throw the ball at 27.7 m/s.
(b) a. The bullet has greater kinetic energy
The kinetic energy of an object is given by
![K=(1)/(2)mv^2](https://img.qammunity.org/2020/formulas/physics/middle-school/c6fs3acuplloc3whu5cpc8ui63cnl7ur39.png)
where m is the mass of the object and v is its speed.
For the bullet, we have:
![K=(1)/(2)(0.0027 kg)(1.50\cdot 10^3 m/s)^2=3037.5 J](https://img.qammunity.org/2020/formulas/physics/college/o9y3kqk41oid7zy3hqr3hhf776o334z4sq.png)
For the ball:
![K=(1)/(2)(0.146 kg)(27.7 m/s)^2=56.0 J](https://img.qammunity.org/2020/formulas/physics/college/7yxigo1wja2bljek6yzghogqqizdg73rsr.png)
So, the bullet has greater kinetic energy.