Engineering Mechanics

Chapter 20 : Projectiles „„„„„ 435

Distance OP
First of all, consider the horizontal motion of the particle. We know that the horizontal
distance OA

x = Horizontal component of velocity × Time

= 100 cos 45° × 3.05 = 100 × 0.707 × 3.05 = 215.6 m ...(i)

Now consider vertical motion of the particle. We know that vertical component of the velocity.

uy= 100 sin 45° = 100 × 0.707 = 70.7 m

and vertical distance AP,^22

11

-. (70.7 3.05) – 9.8 (3.05)
y 22
yut==××gt

= 215.6 – 45.6 = 170 m

∴ Distance OP=+=(215.6)^22 (170) 274.6 m^ Ans.

Example 20.19. A projectile is fired with a velocity of 500 m/s at an elevation of 35°.
Neglecting air friction, find the velocity and direction of the projectile moving after 29 seconds and
30 seconds of firing.

Solution. Given : Velocity of projection (u) = 500 m/s and angle of projection (α) = 35°

Velocity of the projectile after 29 and 30 seconds of firing

We know that velocity of projectile after 29 seconds

=+ugt u^222 –2 (sin )αgt

222

v 39 =+× ××°××⎡⎤⎣⎦(500) (9.8) (29) – (2 500 (sin 35 ) 9.8 29)

=+(250 000 80770) – (1000××0.5736 284.2) m/s

= 409.58 m/s Ans.

Similarly v 30 =+× ××°××⎡⎤⎣⎦(500)^222 (9.8) (30) – (2 500 (sin 35 ) 9.8 30)

=+[][250 000 86440 – 1000 0.5736××294 m/s]

= 409.64 m/s Ans.

Direction of the projectile after 29 and 30 seconds

We know that the angle, which the projectile makes with the horizontal after 29 seconds,

29

sin – (500 sin 35°) – (9.8 × 29)

tan

cos 500 cos 35

ugt

u

α

θ= =

α°

(500 0.5736) – 284.2

0.00635

500 0.8192

×

==

×

or θ= 29 0.36° Ans.

Similarly^30

(500 sin 35 ) – (9.8 30) (500 0.5736) – 294.0

tan

500 cos 35 500 0.8192

°× ×
θ= =
°×
= – 0.0176 or θ= ° 30 –1
Note : Minus sign means that the projectile is moving downwards after reaching the
highest point.