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

 

 

Projectiles - Example 02

 

Projectiles

2. In Fig. 2, a cricket ball is hit at a height h = 1.00 m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.00 s after it is hit and then down past the top of the wall 4.00 s later, at distance D = 50.0 m farther along the wall.
(a) What horizontal distance is traveled by the ball from hit to catch?
What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball’s velocity just after being hit?
(d) How high is the wall?
Fig 2
 

Information given:
ho = 1.00 m
ho = 1.00 m
t1 = 1.00s
t2 = 5.00s
D = (dx2 - dx1) = 50.0 m

From O to P takes 1 s; From P to Q takes 4 s; From Q to f takes 1 s
Time for entire journey = 6 s

(a) Range = dxf

There is no horizontal force and hence no horizontal acceleration.
Horizontal distance coverd in the 4 seconds (from P to Q) = D = 50m
Distance covered in 1 second (Vx) = 50m/4s = 12.5 m-2 --- Eq1
Using x = x o + voxt + ½at2

x = 0 + (12.5 m-2) (6s) + ½(0) (62)

Range = 75m.

(b) Velocity of stone at O = Vector sum of vix and viy

vix = (12.5 m-2) from Eq1

Vetical distance travelled by the ball
Applying y = yo + voyt + ½at2--- Eq2
Now yf = y0 (h = final height of ball = initial height of ball)

So Eq2 becomes h = h + voyt + ½at2--- Eq2

0 = voy (6s) + ½ (-9.81ms-2) (6s) 2

voy = 29.4 ms-1

By Pythagoras:
vi2 = √(vix2 + viy2)
vi2 = √[(12.5 ms-1)2 + (29.4 ms-1)2]

vi = 31.9 ms-1

tan θi = (viy / vix)
tan θf = (29.4 ms-1) / (12.5 ms-1)
θf = tan-1 [(29.4 ms-1) / (12.5 ms-1)]

θf = 67.0°

vi = 31.9 ms-1 at an angle of elevation of 67.0°

(c) Applying y = yo + voyt + ½at2

hw = 1m + (29.4 ms-1) (1s) + ½(-9.81ms-2) (1s)2

hw = 22.5 m.

 

 

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