I think the mark scheme for this question is wrong but I'm not sure.
2. A competitor makes a dive from a high springboard into a diving pool. She leaves the springboard vertically with a
speed of 4 ms-1 upwards. When she leaves the springboard, she is 5 m above the surface of the pool. The diver is modelled
as a particle moving vertically under gravity alone and it is assumed that she does not the springboard as she descends.
Find
a. Her speed when she reaches the surface of the pool, (3)
b. The time taken to reach the surface of the pool. (3)
For question b, surely you have to find the time taken to reach the highest point and then the time taken to come back down again right? I did that and got an answer of 2.4 seconds. In the mark scheme, however, they just use v = u + at using the u as 4 and got t = 1.5s.
This seems wrong to me as they didn't take into consideration the time it took for the diver to go up and then down right?