Page 60 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 4] MOTION IN A VERTICAL PLANE 45

SOLVED PROBLEM 4.12
At what angle should a projectile be fired in order for its range to be a maximum?
The range of a projectile is given by

v 0 2
R = sin 2θ
g
◦
The greatest value the sine function can have is 1. Since sin 90 = 1, the maximum range occurs when
2θ = 90 ◦ θ = 45 ◦
◦
Larger and smaller angles than 45 give shorter ranges. When θ = 45 ,
◦
v 2 0
R max =
g

SOLVED PROBLEM 4.13

(a) What minimum initial velocity must a projectile have to reach a target 90 km away?
(b) What would the time of flight be?
√
2
(a) The maximum range of a projectile of initial velocity v 0 is R = v /g. Solving for v 0 gives v 0 = Rg. Since
0
here
4
R = (90 km)(1000 m/km) = 9 × 10 m
√ 2 √
2
5
we have v 0 = Rg = (9 × 10 m)(9.8 m/s ) = 8.82 × 10 m /s 2
4
5
To find a square root, the power of 10 must be an even number. We therefore write 8.82 × 10 as 88.2 × 10 4
and find that
2
4
2
2
v 0 = 88.2 × 10 m /s = 9.39 × 10 m/s = 939 m/s
(b) Here θ = 45 , corresponding to maximum range, so
◦
◦
2v 0 sin θ (2)(939 m/s)(sin 45 )
T = = = 136 s = 2 min 16 s
g 9.8 m/s 2
SOLVED PROBLEM 4.14
At what two angles above the horizontal can a projectile be fired in order to reach a distance equal to half
its maximum range?
2
2
The maximum range is R max = v /g, and so here R = R max /2 = v /(2g). The smaller angle θ 1 is found as
0 0
follows:
v 0 2 v 2 0
R = sin 2θ 1 =
g 2g
1
sin 2θ 1 = = 0.5
2
2θ 1 = sin −1 0.5
1 −1
θ 1 = sin 0.5 = 15 ◦
2
The larger angle θ 2 is given by
◦
θ 2 = 90 − θ 1 = 90 − 15 = 75 ◦
◦
◦