Assuming that's a squared:

Max height when dh/dt = 0

-32(t-3)=0
t-3=0
t=3

Time 2 secs after max height = 5 secs

-16*(5 - 3)^2 + 150
= -16*4 + 150
= -64 +150
= 86

Answer B

thanks for the explanation but i have a question..
How u can out t = 5 in the equation...
Coz if i calculate height for time one second before and after the max height then these 2 heights should be same.. right?? but if u use the equation in the ques. u will get 2 different heights....
i have a reason for why the 2 height could be different, that i feel is right, but please lemme know what do u think about this and then i'll put my thought

You will have the same height at t=2 and t=4.

t=2
h = -16(2 - 3)^2 + 150
= -16 + 150
= 134

t=4
h = -16 (4 - 3)^2 + 150
= -16 + 150
= 134

oops i,ve taken the wrong equation
lemme correct the equation in post for other viewers
anyways thanks again.....