Answer:
Explanation:
Given
ball is at height of 1.31 m
Horizontal Range=103 m
launch angle[tex]=45 ^{\circ}[/tex]
[tex]Range=\frac{u^2\sin 2\theta }{g}[/tex]
[tex]103=\frac{u^2\sin 90}{9.8}[/tex]
u=31.77 m/s
we know equation of trajectory of Projectile is
[tex]y=x\tan \theta -\frac{gx^2}{2u^2\cos ^2 \theta }[/tex]
here x=93 m
[tex]y=93\times \tan 45-\frac{9.8\times 93^2}{2\times 31.77^2\times \cos ^{2}45}[/tex]
y=9.02 m
Thus at x=93 m ball is at a height of 9.02 above its initial position i.e. 10.33 m above ground
Distance between fence and height of ball=10.33-8.55=1.78 m
