Please help with these two projectile motion problems, thanks

[shaunamariexx]

#1:

Homework Statement

A projectile is launched at an angle of 64.9 degrees with the horizontal (x axis) at some unknown velocity. After 4.6 seconds, the object's angle of trajectory is 40.3 degrees above the horizontal. What is the object's vertical displacement at this time in meters? (Hint: first find the magnitude of the initial velocity using the angle of trajectory (draw a triangle and relate components) and the equation for the final velocity in the y direction.The attempt at a solution
I first tried to find the magnitude of the initial velocity using the trajectory but I had some trouble! And then I know there is an equation for the final velocity in the y direction to help with the answer.

#2:
Homework Statement
A projectile is lauched at some speed at an angle of 62 degrees and travels a range of 220 meters. The launch angle is then changed, but the projectile is fired at the same speed as the original case and also travels the same range of 220 meters. Find the maximum vertical displacement at 62 degrees, (ymax1), and find the maximum vertical displacement for the second angle (ymax2), and then find the difference between them in meters

The attempt at a solution
I know after I find the ymax values to subtract ymax2 from ymax 1

 

[cnh1995]

A projectile is launched at an angle of 64.9 degrees with the horizontal (x axis) at some unknown velocity. After 4.6 seconds, the object's angle of trajectory is 40.3 degrees above the horizontal. What is the object's vertical displacement at this time in meters? (Hint: first find the magnitude of the initial velocity using the angle of trajectory (draw a triangle and relate components) and the equation for the final velocity in the y direction.

Let u_y= initial vertical component of velocity and v_y=final vertical velocity.
Horizontal component of velocity i.e. v_x is same throughout the journey.
tan(64.9°)=u_y/v_x
tan(40.3°)=v_y/v_x
Taking ratio,
u_y/v_y=2.5172
∴u_y=2.5172v_y-------(1)
Also,
v_y=u_y-gt
From 1,
v_y=2.5172v_y-gt
∴1.5172v_y=gt
∴v_y=gt/1.5172
∴v_y=29.74m/s
u_y=74.86m/s
Also, vertical displacement
s=(u_y²-v_y²)/2g
∴s=240.80m

 

[cnh1995]

A projectile is lauched at some speed at an angle of 62 degrees and travels a range of 220 meters. The launch angle is then changed, but the projectile is fired at the same speed as the original case and also travels the same range of 220 meters. Find the maximum vertical displacement at 62 degrees, (ymax1), and find the maximum vertical displacement for the second angle (ymax2), and then find the difference between them in meters

For having the same range, the angles should be complementary i.e. θ1+θ2=90°.
(Because R=v²sin(2θ)/g
and if θ1+θ2=90°, sin(2θ₁)=sin(2θ₂)).
So, the other angle of projection is 28°.
Rest of the calculations are pretty straightforward.