Answer:20.82 m
Explanation:
Given
distance between wall and launcher=5 m
initial velocity=30 m/s
Launching angle[tex]=45^{\circ}[/tex]
Height of wall=12 m
maximum height by ball
[tex]h_{max}=\frac{u^2sin^2\theta }{2g}[/tex]
[tex]h_{max}=\frac{30^2sin^{2}45}{2\times 9.8}[/tex]
h=22.95 m
[tex]y=xtan\theta -\frac{gx^2}{2u^2cos^\theta }[/tex]
y=4.72 m
so ball will strike with wall
and for perfect restitution final velocity of ball will be same as the horizontal velocity before its impact, only direction will be opposite and vertical velocity will be zero
thus it seems as if someone throw the ball with horizontal velocity of 30cos45 from a height of 4.72
Time required to cover 4.72 m
[tex]t=\sqrt{\frac{2h}{g}}[/tex]
[tex]t=\sqrt{\frac{9.45}{9.8}}[/tex]
t=0.981 s
Horizontal distance traveled in this time
[tex]R=ucos45\times t[/tex]
[tex]R=30\times cos45\times 0.981[/tex]
R=20.82 m
