UNIT 4
PHYS 2001
Example 01
Example 02
Example 03
Example 04
Example 05
Example 06
Example 07

 

 

Projectiles - Example 01

 

Projectiles

1. In Fig. 1, a stone is projected at a cliff of height h with an initial speed of 42.0 ms-1 directed at angle θo = 60.0° above the horizontal. The stone strikes at A, 5.50 s after launching.
Find (a) the height h of the cliff,
(b) the velocity of the stone just before impact at A, and
(c) the maximum height H reached above the ground.
Fig 1
 

Information given:|
θo = 60.0°
vo= 42.0 ms-1
t1 = 5.50s
yo= 0.00 m.

(a) Using the equation of motion in the vertical plane:
y = yo + voyt + ½at2 --- Eq1
and voy = vo sin 60.0° and a = g = -9.80 ms-2 and t = 5.50s

So Eq 1 becomes: y = 0.00 + (42.0 ms-1 sin 60.0°) (5.50s) + ½ (-9.80 ms-2) (5.50s)2
y = 52.0 m

Hence h = 52.0 m

(b) Velocity of stone at A = Vector sum of vfx and vfy

Horizontal acceleration = 0.00 ms-2

Horizontal velocity: vfx = vox + (0.00 ms-2) (t)
vfx = (vocos 60.0°) + (0.00 ms-2 )(5.50s)
vfx = 21.0 ms-1

Vertical acceleration = g ms-2

Vertical velocity: vy = voy + (g) (t)
vfy = (vo sin 60.0°) + (-9.80 ms-2) (5.50s)
vfy = -17.5 ms-1

By Pythagoras:
vf2 = √(vfx2 + vfy2)
vf2 = √[( 21.0 ms-1)2 + (-17.5 ms-1)2]

vf = 27.3 ms-1

tan θf = (vy / vx)
tan θf = (-17.5 ms-1) / (21.0 ms-1)
θf = 39.8°

vo = 27.3 ms-1 at and angle of depression of 39.8°

(c) At maximum height vy = 0.00 ms-1

From vy2 = voy2 + 2gh

02 = [(42.0 ms-1) sin 60.0°)]2 + 2 X (-9.80 ms-2) H

H = 67.5 m.

 

 

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Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.