Question

A small racquet ball launcher is set up 5 meters from a 12 meter tall wall. It launches a racket ball at 30 m/s at an angle of 45 degrees. Determine where the racket ball lands relative to the wall if there is no air resistance and the racquet ball has perfect restitution with the wall.

Answer 1

Answer:20.82 m

Explanation:

Given

distance between wall and launcher=5 m

initial velocity=30 m/s

Launching angle$=45^{\circ}$

Height of wall=12 m

maximum height by ball

$h_{max}=\frac{u^2sin^2\theta }{2g}$

$h_{max}=\frac{30^2sin^{2}45}{2\times 9.8}$

h=22.95 m

$y=xtan\theta -\frac{gx^2}{2u^2cos^\theta }$

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

$t=\sqrt{\frac{2h}{g}}$

$t=\sqrt{\frac{9.45}{9.8}}$

t=0.981 s

Horizontal distance traveled in this time

$R=ucos45\times t$

$R=30\times cos45\times 0.981$

R=20.82 m