Motion in a Plane

To analyze the problem, consider two objects:
First Object: Maximizing Range
To achieve maximum range, the object must be projected at a 45∘ angle.
Using the formula for maximum height H1 when launched at an angle θ1=45∘ :
H1=‌

u12sin‌2θ1

2g

=‌

u12×1

2g×2

=‌

u12

4g

Second Object: Maximizing Height
To achieve maximum height, the object should be projected vertically at a 90∘ angle.
The formula for the maximum height H2 with a launch angle θ2=90∘ :
H2=‌

u22sin‌290∘

2g

=‌

u22

2g

Equating Heights
Since both objects reach the same maximum height:
H1=H2
From equations (i) and (ii):
‌

u12

4g

=‌

u22

2g

Simplifying, we find:
u12=2u22
Taking the square root:
u1=√2u2
Thus, the ratio of their initial velocities is:
‌

u1

u2

=‌

√2

1

Therefore, the ratio of initial velocities is √2:1.