Answer:
The second projectile is [tex]\sqrt{2}[/tex] times faster than the first projectile.
Explanation:
Mass of first projectile = m
height [tex]h_{1}[/tex] = 2.6 cm
Height [tex]H_{2}[/tex] = 5.2 cm
Let velocity of first projectile = V
Velocity of first projectile = V'
From conservation of momentum principal
m V = (m + m) V'
[tex]V = \frac{m +m}{m} V'[/tex]
V' = [tex]\sqrt{2gH}[/tex]
[tex]V = \frac{m +m}{m} \sqrt{2gH}[/tex]
Therefore velocity is directly proportional to the height.
[tex]\frac{V_{2} }{V_{1} } =\frac{ \sqrt{H_{2} } }\sqrt{H_{1}[/tex]
Since [tex]H_{2} = 2 H_{1}[/tex]
[tex]\frac{V_{2} }{V_{1} } = \sqrt{2}[/tex]
Therefore the second projectile is [tex]\sqrt{2}[/tex] times faster than the first projectile.
