A woman throws a ball at a vertical wall d = 5.4 m away. The ball is h = 2.7 m a

Question

A woman throws a ball at a vertical wall d = 5.4 m away. The ball is h = 2.7 m above ground when it leaves the woman's hand with an initial velocity of 10 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?

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

I have already found b & c and they were correct.

1.98 s

(c) Where did the ball hit the wall?

5.24 m

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

Explanation / Answer

vi = 10 m/s
Angle = 45°
d = 5.4 m
h = 2.7 m
Horizontal component of velocity = 10*cos(45) = 7.1 m/s
Vertical  component of velocity = 10*cos(45) = 7.1 m/s
Horizontal acceleration = 0
Vertical acceleration = 9.8 m/s^2

time taken to reach the wall, t = 5.4/7.1 = 0.76 s
Vertical velocity of ball when it hits the wall, v = u + a*t
v = 7.1 - 9.8*0.76
v = -0.348

s = u*t + 1/2at^2
s = 7.1*0.76 - 1/2*9.8*0.76^2
s = 2.56 m
It hit's the ball. s = 2.56 + 2.7 = 5.26 m

Now after hitting the wall,
Horizontal component of velocity, = 10 * cos(45) = 7.1 m/s
Vertical component = 0.348 m/s

Now time taken by ball to reach ground,
s = u*t + 1/2at^2
5.26 = 0.348 * t + 1/2*9.8*t^2
t = 1.0 s

(a)
Horizontal Distance travelled by ball in this time, = 7.1 * 1.0 = 7.1 m
So ball hit's the ground 7.1 m from wall.

(d)
How long was the ball in the air after it left the wall ?
t = 1.0 s

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