Answer:
a) 3.56m/s
b) 1.73m
Step-by-step explanation:
We have to treat this as a parabolic motion problem:
we will use the next formulas:
![V=Vo+a*t\\Vy^2=Vyo^2+2*a*Y\\X=Vox*t](https://img.qammunity.org/2020/formulas/physics/college/6fiywrb5smwm4afrstcih0f3u64cn3w7js.png)
we first have to calculate the initial velocity of the basketball player:
![Vy^2=Vyo^2+2*a*Y\\0^2=Vyo^2+2*(-9.8)*(0.650)\\Vyo=√(2*9.8*0.650) \\Vyo=3.56 m/s](https://img.qammunity.org/2020/formulas/physics/college/rv9n4w3w2g18hlwnjtmvnk39bp3rk9j2bv.png)
the final velocity is zero when he reaches the maximun height.
To answer the second part we need to obtain the time to reach the maximun height, so:
![V=Vo+a*t\\\\0=3.56+(-9.8)*t\\t=0.36 seconds](https://img.qammunity.org/2020/formulas/physics/college/6o4p9vdrq5rup2eqiey73v2s4pd29wgvk6.png)
Now having that time, let's find the distance on the X axis, the X axis behaves as constant velocity movement, so:
![X=Vox*t\\X=4.80*0.36\\X=1.73m](https://img.qammunity.org/2020/formulas/physics/college/tichgtr40ut55ssf7d96j9vwd46btjdexu.png)