1.3 Solutions
31
In the same time, the y-coordinate at D is given by
y = H = u y t −
1
2
gt
2
= u(sin α)t −
1
2
gt
2
H = u
h
s
s
u
−
1
2
gt
2
= h −
1
2
gt
2
or h − H =
1
2
gt
2
∴ t =
2(h − H )
g
But the quantity (h–H ) represents the height through which the monkey drops
from the tree and the right-hand side of the last equation gives the time for a
free fall. Therefore, the bullet would hit the monkey independent of the bullet’s
initial velocity.
1.33 R =
u 2 sin 2α
g
, h =
u 2 sin
2
α
g
, T =
2u sin α
g
(a) h
R
=
1
4
tan α → tan α =
4h
R
(b) h
T 2 =
g
8
→ h =
gT 2
8
1.34 (i) T =
2u sin α
g
=
2 × 800 sin 60 ◦
9.8
= 141.4 s
(ii) R =
u 2 sin 2α
g
=
(800) 2 sin(2 × 60)
9.8
= 5.6568 × 10 4 m = 56.57 km
(iii) Time to reach maximum height =
1
2 T =
1
2 × 141.4 = 707 s
(iv) x = (u cos α)t
(1)
y = (u sin α)t −
1
2
gt
2
(2)
Eliminating t between (1) and (2) and simplifying
y = x tan α −
1
2
gx 2
u 2 cos 2 α
(3)
which is of the form y = bx +cx 2 , with b = tan α and c = −
1
2
g
u 2 cos 2 α
.
1.35 (i) T =
u sin α
g
=
350 sin 55 ◦
9.8
= 29.25 s