A woman throws a ball at a vertical wall d = 4.6 m away. The ball is h = 2.3 m a

Question

A woman throws a ball at a vertical wall d = 4.6 m away. The ball is h = 2.3 m above ground when it leaves the woman's hand with an initial velocity of 11 m/s at 45degree. 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.) Where does the ball hit the ground? m (away from the wall) How long was the ball in the air before it hit the wall? s Where did the ball hit the wall? m (above the ground) How long was the ball in the air after it left the wall? s

Explanation / Answer

height h=2.3 m

d=4.6 m away from the wall

initial velocity u=11 m/sec

angle theta=45 degrees,

horizontal velocity, vx=v*cos(theta)

vx=11*sin(45)=7.78 m/sec

vertical velocity vy=v*sin(theta)

vy=11*sin(45) =7.78 m/sec

and

time taken by the ball to reach the wall is,

ti=d/vx

ti=4.6/7.78

ti=0.59 sec

vertical elevation before hitting the wall is,,

y=h+(vy*ti)-(1/2*g*ti^2)

y=2.3+(7.78*0.59)-(1/2*9.8*0.59^2)

y=5.18 m

after impact with the wall,

y=1/2*g*tf^2

5.18=1/2*9.8*tf^2

==>

tf=1.03 sec

now,

after impact,

the range of the ball is R=vx*tf

R=7.78*1.03

R=8.01 m

a)

the ball hit the ground away from the wall is ,

R=8.01 m

which is (8.01-4.6)=3.41 m behind the position of the woman

b)

time taken by the ball before hit the wall is,

ti=0.59 sec =0.6 sec

c)

the vertical height of the ball

y=5.18 m from the ground

d)

time taken by the ball in air before hit the ground is,

Tf=1.03 sec

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