Aball is thrown upward with an initial velocity of 20 m/s. (a) how high above the thrower’s hand is the ball after 2 seconds? (b) how long is the ball in the air? a ball is thrown upward with an initial velocity of 20 m/s. (c) what is the ball’s instantaneous velocity after 1 second? (d) what is the ball’s displacement after 1 second? (e) what is the ball’s average velocity between t = 1 and t = 4 seconds?

The ball goes up vertically only with positive velocity. It reaches a maximum height where velocity is zero then the ball falls down with negative velocity.

Best place to start is by finding the time it takes to reach maximum height.
v = u + at
0 = 20 - 9.8t
9.8t = 20
t = 20/9.8
t = 2 seconds
So it takes 2 sec to reach max height then it will be 4 sec to return to it's starting point.

A. find the height at 2 sec.
Δy = ut + (1/2)at²
Δy = 20(2) + (1/2)(-9.8)(2)²
Δy = 40 - 19.6
Δy = 20.4 m

B. we found that the ball is in the air for 4 seconds;
2 sec going up , 2 sec going down.

C. Instantaneous velocity at t = 1 sec ; will be going up +direction
v = u + at
v = 20 - 9.8(1)
v = 10.2 m/s upward

D. displacement, Δy at t = 1 sec
Δy = 20(1) -  4.9(1)²
Δy = 20 - 4.9
Δy = 15.1 m

E. Average velocity = Δy / Δt
We found that at t = 1 sec the ball is at 15.1 m and at t = 4 sec the ball has returned to starting point, 0 m.
Δy =  (0 - 15.1) = -15.1 m.
Δt = (4 - 1) = 3 sec
Average velocity = -15.1/3 =  -5 m/s

My answer is opptionc, just check with anyone else