34
1 Kinematics and Statics
Fig. 1.22
u x = u cos α = u cos 45
◦
=
u
√
2
u y = u sin α = u sin 45
◦
=
u
√
2
When the ball hits the wall, y = x tan α −
1
2
gx 2
u 2 cos 2 α
Using y = H , x = d and α = 45 ◦
H = d
1 −
gd
u 2
(1)
If the collision of the ball with the wall is perfectly elastic then at P, the
horizontal component of the velocity (u
x ) will be reversed, the magnitude
remaining constant, while both the direction and magnitude of the vertical
component v
y are unaltered. If the time taken for the ball to bounce back from
P to A is t and the range BA = R
y = v
y t −
1
2
gt
2
(2)
Using t =
R
u cos 45 ◦ =
√
2
R
u
(3)
y = −(H + h)
(4)
v
y t = u sin 45
◦
− g
d
u cos 45 ◦ =
u
√
2
−
√
2
gd
u
(5)
Using (3), (4) and (5) in (2), we get a quadratic equation in R which has the
acceptable solution
R =
u 2
2g
+
u 2
4g 2 + H + h
1 Kinematics and Statics
Fig. 1.22
u x = u cos α = u cos 45
◦
=
u
√
2
u y = u sin α = u sin 45
◦
=
u
√
2
When the ball hits the wall, y = x tan α −
1
2
gx 2
u 2 cos 2 α
Using y = H , x = d and α = 45 ◦
H = d
1 −
gd
u 2
(1)
If the collision of the ball with the wall is perfectly elastic then at P, the
horizontal component of the velocity (u
x ) will be reversed, the magnitude
remaining constant, while both the direction and magnitude of the vertical
component v
y are unaltered. If the time taken for the ball to bounce back from
P to A is t and the range BA = R
y = v
y t −
1
2
gt
2
(2)
Using t =
R
u cos 45 ◦ =
√
2
R
u
(3)
y = −(H + h)
(4)
v
y t = u sin 45
◦
− g
d
u cos 45 ◦ =
u
√
2
−
√
2
gd
u
(5)
Using (3), (4) and (5) in (2), we get a quadratic equation in R which has the
acceptable solution
R =
u 2
2g
+
u 2
4g 2 + H + h