Answer:
(a) 19.6 ms⁻¹
(b) 2 s
(c) 9.8 ms⁻¹
(d) 4 s
Step-by-step explanation:
<u>Constant Acceleration Equations (SUVAT)</u>

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.
Acceleration due to gravity = 9.8 ms⁻².
<h3><u>Part (a)</u></h3>
When the ball reaches its maximum height, its velocity will momentarily be zero.
<u>Given values</u> (taking up as positive):


Therefore, the initial speed is 19.6 ms⁻¹.
<h3><u>Part (b)</u></h3>
Using the same values as for part (a):

Therefore, the time taken to reach the highest point is 2 seconds.
<h3><u>Part (c)</u></h3>
As the ball reaches its maximum height at 2 seconds, one second before this time is 1 s.
<u>Given values</u> (taking up as positive):


The velocity of the ball one second before it reaches its maximum height is the <u>same</u> as the velocity one second after.
<u>Proof</u>
When the ball reaches its maximum height, its velocity is zero.
Therefore, the values for the downwards journey (from when it reaches its maximum height):

(acceleration is now positive as we are taking ↓ as positive).

Therefore, the velocity of the ball one second before <u>and</u> one second after it reaches the maximum height is 9.8 ms⁻¹.
<h3><u>Part (d)</u></h3>
From part (a) we know that the time taken to reach the highest point is 2 seconds. Therefore, the time taken by the ball to travel from the highest point to its original position will also be 2 seconds.
Therefore, the total time taken by the ball to return to its original position after it is thrown upwards is 4 seconds.