Given that,
Height =1.5 m
Angle = 45°
We need to find the greater speed of the ball
Using conservation of energy
![P.E_(i)+K.E_(f)=P.E_(f)+K.E_(f)](https://img.qammunity.org/2021/formulas/physics/college/5ev7nu2vuusfdi5fh9ysem0prlq02onkmy.png)
![mgh+(1)/(2)mv_(i)^2=mgh+(1)/(2)mv_(f)^2](https://img.qammunity.org/2021/formulas/physics/college/3srajsq9eep0deaubn1wc5nbqeiorj9fvv.png)
Here, initial velocity and final potential energy is zero.
![mgh=(1)/(2)mv_(f)^2](https://img.qammunity.org/2021/formulas/physics/college/85s3ch5y28rn65k3et6ko6ry6q4kszc8bs.png)
Put the value into the formula
![9.8*1.5=(1)/(2)v_(f)^2](https://img.qammunity.org/2021/formulas/physics/college/add75rzeshldnvfe17zjhdv9k5xlvbvbju.png)
![v_(f)^2=2*9.8*1.5](https://img.qammunity.org/2021/formulas/physics/college/t68n9xb1tqrpd71jybaekhhk24wd90u21m.png)
![v_(f)=√(2*9.8*1.5)](https://img.qammunity.org/2021/formulas/physics/college/rpexldw76u608r0k7y6zjaa620atg2f7x7.png)
![v_(f)=5.42\ m/s](https://img.qammunity.org/2021/formulas/physics/college/6p9okk0a4djasss5kaulk72ldrh0wndi0l.png)
Hence, the greater speed of the ball is 5.42 m/s.