A ball is thrown from a height of 65m with initial velocity in y direction of 41ms. How long does it take the ball to reach the ground?

I'll ASSUME that you mean 41 m/s. (not ms)

There are two potential solutions because you have not oriented the reference frame.

IF UP is the positive direction and ground level is the origin.

s = s₀ + v₀t + ½at²

0 = 65 + 45t + ½(-9.8)t²

0 = 65 + 45t - 4.9t²

quadratic formula

t = (-45 ±√(45² - 4(-4.9)(65))) / (2(-4.9))

t = -1.26907... which we ignore as it occurs before the ball is thrown 

                               (But note the magnitude)

or

t = 10.4527... s

which, when rounded to the two significant digits of the question numerals becomes

t ≈ 10 s  ◄

IF DOWN is the positive direction and ground level is the origin.

the equation becomes

0 = -65 + 45t + ½(9.8)t²

0 = -65 + 45t + 4.9t²

quadratic formula

t = (-45 ±√(45² - 4(4.9)(-65))) / (2(4.9))

t = -10.4527... which we ignore for the same reason as before.

or

t = 1.26907... s

t ≈ 1.3 s  ◄

My GUESS would be the first solution with more generally ASSUMED conventions, but it would not have to be...the reference frame is important.

Note that in each case, the negative answer is the time needed for the mass to get to the throwing point if it were launched from ground level at the proper velocity to arrive at the throwing point altitude with the indicated initial velocity.

I hope this helps.

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h = h0 + v0*t - (1/2)gt^2.

Here h = 0, h0 = 65 m, v0 = 41 m/s, g = 9.8 m/s^2

and you get a quadratic equation with "t" as the unknown.

Use quadratic formula.

Bill Russell makes a legitimate point, but I am assuming the ball is thrown straight upward.

My guess at an answer is around 9 or 10 seconds,

because it takes around 4 seconds for the ball to lose its upward velocity (41/10 = 4), and another 4 seconds to get back to the 65 position, and another 1+ second to get down to the ground.  It helps to have a preliminary guess in case you make a mistake with the quadratic formula!

41 ms ? 41 milliseconds?  assuming m/s

but you need the angle to solve this problem.