Part A)
when ball will reach to highest point then it's speed will become zero
so we can use kinematics to find the time
![v_f = v_i + at](https://img.qammunity.org/2020/formulas/physics/middle-school/1cset2hbhcdxo7q34bqods06bywswu8y3w.png)
![0 = 21.9 + (-9.8) t](https://img.qammunity.org/2020/formulas/physics/middle-school/6ieac3p43ma55jbf237b16gjxjqps88fla.png)
![0 = 21.9 - 9.8 t](https://img.qammunity.org/2020/formulas/physics/middle-school/cglx0dsusmau3exxvwq58qm0k9j44f0pgu.png)
![t = 2.23 s](https://img.qammunity.org/2020/formulas/physics/middle-school/9pasy6hvi4j5srviy8sfgzmuypw071128c.png)
Part b)
for finding the maximum height we can use another kinematics equation
![v_f^2 - v_i^2 = 2ad](https://img.qammunity.org/2020/formulas/physics/middle-school/49zbwj6kktn1fcojghv6c4pfmbqzbqgva7.png)
![0 - 21.9^2 = 2(-9.8)(H)](https://img.qammunity.org/2020/formulas/physics/middle-school/z4jslojbkkokzyu2xcd1wiivxe64tmu413.png)
![H = (21.9^2)/(19.6) = 1.12 m](https://img.qammunity.org/2020/formulas/physics/middle-school/383qgn060sgipnrn94vle16y11fu2upo8w.png)
so it will rise to 1.12 m from the point of projection
Part C)
Ball will take double the time which it take to reach the top point.
So here the time to reach the top is 2.23 s
so time taken by the ball to reach at same point after projection is given as
![t = 2(2.23) = 4.46 s](https://img.qammunity.org/2020/formulas/physics/middle-school/q9xbhxslo3863blbb5flafst04d9ku6e0l.png)
Since ball have reached to same point so the final velocity must be same as initial velocity
so we have
downwards
Part d)
when ball reached to the bottom
displacement of ball = -51.6 m
![a = -9.8 m/s^2](https://img.qammunity.org/2020/formulas/physics/middle-school/vkpxtimpvw776aeywe9lp8fwpopfx1wyg6.png)
![v_i = 21.9 m/s](https://img.qammunity.org/2020/formulas/physics/middle-school/jd1gkm7m7glx36dhrrvtc1vk2dy9tmce69.png)
now by kinematics we have
![d = v_i t + (1)/(2)at^2](https://img.qammunity.org/2020/formulas/physics/middle-school/u8fu4rcfes4n1mmseb9xi0sm2xhtbmhfhi.png)
![-51.6 = (21.9)t + (1)/(2)(-9.8)t^2](https://img.qammunity.org/2020/formulas/physics/middle-school/qlg5ye346dkzs0tcurif353stnpn3awh8k.png)
![4.9 t^2 - 21.9 t - 51.6 = 0](https://img.qammunity.org/2020/formulas/physics/middle-school/p28b5hiwrbwd6uuabz8yz5l38ifkqm8nv1.png)
by solving above equation we have
![t = 6.2 s](https://img.qammunity.org/2020/formulas/physics/middle-school/hha2r45c2ihq6rz4hw0jbt2roan8741snn.png)
now for the velocity at that instant we have
![v_f = v_i + at](https://img.qammunity.org/2020/formulas/physics/middle-school/1cset2hbhcdxo7q34bqods06bywswu8y3w.png)
![v_f = 21.9 - (9.8) (6.2)](https://img.qammunity.org/2020/formulas/physics/middle-school/yt3r7rab9y1wd28n8kqehiikylfjhle2jy.png)
![v_f = -38.6 m/s](https://img.qammunity.org/2020/formulas/physics/middle-school/p8erzo3rba3dcgnq8sp8zo2psk60lu54nk.png)
so its velocity is 38.6 m/s downwards
Part e)
for the position of ball at t = 5.35 s we can use
![d = v_i t + (1)/(2)at^2](https://img.qammunity.org/2020/formulas/physics/middle-school/u8fu4rcfes4n1mmseb9xi0sm2xhtbmhfhi.png)
![d = 21.9(5.35) + (1)/(2)(-9.8)(5.35)^2](https://img.qammunity.org/2020/formulas/physics/middle-school/ksrpwb2y26rs536bi3g9spj02psxffzrn1.png)
![d = -23.1 m](https://img.qammunity.org/2020/formulas/physics/middle-school/w1lz5y3yyxobrh1gbry9z4kv3dx7z61yys.png)
so it is 23.1 m below the initial position from which it is thrown
now for the velocity we can say
![v_f = v_i + at](https://img.qammunity.org/2020/formulas/physics/middle-school/1cset2hbhcdxo7q34bqods06bywswu8y3w.png)
![v_f = 21.9 + (-9.8)(5.35)](https://img.qammunity.org/2020/formulas/physics/middle-school/bmjrurv9f6i5qpacoovn4xkpxb7gyu8bzn.png)
![v_f = -30.53 m/s](https://img.qammunity.org/2020/formulas/physics/middle-school/dj0t5lmzwd2cgpeiw89yu4gopnucryshr6.png)
so it will be moving downwards with speed 30.53 m/s