Answer:[tex]27^{\circ}[/tex]
Explanation:
Given
Initial velocity of both snowball is 29.3 m/s
first snowball launch angle[tex]=63^{\circ}[/tex]
Considering motion of snowball to be projectile
range is given by
[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
[tex]R=\frac{29.3^2\sin 126}{9.8}[/tex]
R=70.87 m-----1
If second snowball is thrown at an angle of \phi
[tex]R=\frac{u^2\sin 2\phi }{g}[/tex]
[tex]R=\frac{29.3^2\sin 2\phi }{9.8}[/tex]------2
[tex]70.87=87.601\sin 2\phi [/tex]
[tex]0.809=\sin \phi [/tex]
[tex]2\phi can be 53.99^{\circ}[/tex]
or [tex]180-2\phi =53.99^{\circ}[/tex]
Thus [tex]\phi =26.995 \approx 27^{\circ}[/tex]