A woman throws a ball at a vertical wall d = 3.6 m away. The ball is h = 1.8 m a

Question

A woman throws a ball at a vertical wall d = 3.6 m away. The ball is h = 1.8 m above ground when it leaves the woman's hand with an initial velocity of 11 m/s at 45°. When the ball hits the wall, the horizontal component of its velocity is reversed; the vertical component remains unchanged. (Ignore any effects due to air resistance.)
(a) Where does the ball hit the ground?
m (away from the wall)

(b) How long was the ball in the air before it hit the wall?
s

(c) Where did the ball hit the wall?
m (above the ground)

(d) How long was the ball in the air after it left the wall?

Explanation / Answer

d = 3.6 m

h = 1.8 m

v=11 m/s

= 45°

Vertical component of velocity =vv= v Sin = 7.78 m/s

Horizontal component of velocity =vh= v Cos = 7.78 m/s

(b) time taken by ball to hit the vall t1 = d/vh= 0.463 s  

vertical distance covered in this time S= vvt1+(1/2)at12 where   a = -g

                                            S = 2.55 m

(c) The ball hit the wall at a height of H = S+1.8 m = 4.35 m from ground  

vertical velocity when ball hit the wall Vv=vv+at1 where a=-g

                                                          Vv = 3.24 m/s

After hitting wall the direction is reversed but vertical velocity is the same

H= Vvt+(1/2)at2 (a=g)

    4.35 = 3.24* t+ 0.5* 9.8 *t2   

Solving the quadratic equation

t= 0.67 s

(d)   time ball spend in the air after it left the wall = 0.67 s  

horizontal velocity when ball hits the wall Vh= vh+at  

along horizontal direction a=0

So Vh= vh =7.78 m/s which got reversed aftet hitting at wall

(a) total horizontal distance it travels after hitting the wall =R =Vh* t = 5.21 m

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