30
1 Kinematics and Statics
1.31 The equation to the parabolic path can be written as
y = ax − bx
2
(1)
with a = tan θ ; b =
g
2u 2 cos 2 θ
(2)
Taking the point of projection as the origin, the coordinates of the two openings in the windows are (5, 5) and (11, 7), respectively. Using these coordinates in (1) we get the equations
5 = 5a − 25b
(3)
7 = 11a − 121b
(4)
with the solutions, a = 1.303 and b = 0.0606. Using these values in (2), we
find θ = 52.5 ◦ and u = 14.8 m/s.
1.32 Let the rifle be fixed at A and point in the direction AB at an angle α with the
horizontal, the monkey sitting on the tree top at B at height h, Fig. 1.21. The
bullet follows the parabolic path and reaches point D, at height H , in time t.
Fig. 1.21
The horizontal and initial vertical components of velocity of bullet are
u x = u cos α; u y = u sin α
Let the bullet reach the point D, vertically below B in time t, the coordinates
of D being (d, H ). As the horizontal component of velocity is constant
d = u x t = (u cos α)t =
udt
s
where s = AB:
t =
s
u
The vertical component of velocity is reduced due to gravity.