Angle of a projectile

Find the angle of projection at which the horizontal range and the maximum height of a projectile are equal.

So, this is my work, but I got the answer wrong, and I don't know what I did wrong. Can you please point out for me ?

Take upwards as positive, Vi is initial velocity, xi and yi will be the initial position of the projectile, A will be the angle of the projectile with horizontal.
When the projectile is at its max. height, its Vy = 0, so we have
(Vi sinA) - gt = 0 ---------> (Vi sinA) = gt (1)

Horizontal range of the projectile is
x = xi + (Vi cosA)t
x = (Vi cosA)t

Max. height is
y max = yi + (Vi sinA)t - (1/2) gt^2
y max = (Vi sinA)t - (1/2) gt^2
Also, we are given that x = y max. so
(Vi cosA)t = (Vi sinA)t - (1/2) gt^2

since t will be not equal to 0, we can divide t
(Vi cosA) = (Vi sinA) -(1/2) gt (2)

From (1), we can replace (Vi sinA) = gt, so (2) will become
(Vi cosA) = (Vi sinA) - (1/2)(Vi sinA)

(Vi cosA) = (1/2)(Vi sinA)

sinA / cosA = 2

tan A = 2

A = 63.4 degrees

However, the answer is tan A = 4, and A = 76 degrees.....

The time you use for y max is the half the time for the range.
Horizontal range of the projectile is
x = (Vi cosA) (2t) Note the number 2 in this equation.

Now proceed and get the correct value.
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The formula for range R = u^2 sin 2θ /g and
The formula for height is H = u^2 (sinθ)^2 /2g
If R = H

u^2 sin 2θ /g = u^2 (sinθ)^2/ 2g

Canceling u^2/ g on both sides
2 sin θ* cos θ = (sinθ) ^2 /2
tan θ = 4
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