Physics Projectile Motion Problems :( (1 Viewer)

[coeyz]

A football is kicked off the ground at an angle of 30 degree to the horizontal. It is initially moving at 23.0 ms-1. Calculate :

a) The vertical velocity of the ball after 2.00 s (ANS* 6.1ms-1 down)
b) The velocity of the ball after 2.00 s (ANS* 21.5ms-1 22.1degree to horizontal)
c) The height of the ball after 2.00s (ANS* 3.4m)
d) How long into the flight the ball reaches its maximum height (ANS* 1.175)
e) The range of the kick. (ANS* 46.7m)


THANKS A LOT :*(

 

[tris123]

a)
vertical component: 23sin30 = 11.5

Vy=Uy+AyT
Vy = 11.5 x (-9.8 x 2)
Vy = -8.1 (8.1ms-1 downwards)

b)
Horizontal component: 23cos30 = 19.92
Vector vertical + horizontal:
V^2= 8.1^2 + 19.92^2
V= 21.5 ms-1

tan-1 8.1/19.92
= 22.1 degree to horizontal

c)
change in Y = UyT + 1/2AT^2
= (8.1x2) + (1/2 x -9.8) x (2^2)
= 3.4 metres

d)
maximum height:
Vy = Uy^2 + 2Ay x change in Y
0 = 8.1^2 + (2x-9.8) x (change in Y)
change in Y = 3.3m
thus maximum height = 3.3m

time taken to reach maximum height:
vsin angle/9.8

23sin30/9.8

= 1.17 s

e)
Horizontal component = 19.92 ms-1
Time taken for football to come to rest = 1.7 x 2 (question before)
= 2.35

Horizontal component is constant thus: v = u = 19.92 ms-1

change in X = U x T
X = 19.92 x 2.35
X = 46.8m
(close enough)

-----------------------
hope that helps

 

[izah123]

tris123 is so cool

 

[coeyz]

oooo i love you!!!
thankyou so muchhhh

 

[coeyz]

i dont understand here

a)
vertical component: 23sin30 = 11.5

Vy=Uy+AyT
Vy = 11.5 x (-9.8 x 2)
Vy = -8.1 (8.1ms-1 downwards)

b)
Horizontal component: 23cos30 = 19.92
Vector vertical + horizontal:
V^2= 8.1^2 + 19.92^2
V= 21.5 ms-1

tan-1 8.1/19.92
= 22.1 degree to horizontal


In b, to calculate the velocity,
it should use vertical + horizontal component,
or vertical velocity + horizontal velocity?
why here you use Vy^2+ horizontal component^2?

 

[coeyz]

and here,..

d)
maximum height:
Vy = Uy^2 + 2Ay x change in Y
0 = 8.1^2 + (2x-9.8) x (change in Y)
change in Y = 3.3m
thus maximum height = 3.3m

time taken to reach maximum height:
vsin angle/9.8

23sin30/9.8

= 1.17 s

--
how can u get the Uy= 8.1 ?
i thought 8.1 is the Vy .. (ans of question a)

 

[tris123]

you add the vertical velocity + horizontal velocity using pythags theory (c^2 = a^2 + b^2) to determine the final velocity.

horizontal velocity = 23cos30
vertical velocity = 23syn30

if it helps draw a right angled triangle and label the 2 speeds then calculate the hypotenuse to find the answer

is that what your asking?

 

[tris123]

for Q d) it doesn't matter for that particular Q you will come out with same answer, your right tho i made a mistake

 

[coeyz]

you add the vertical velocity + horizontal velocity using pythags theory (c^2 = a^2 + b^2) to determine the final velocity.

horizontal velocity = 23cos30
vertical velocity = 23syn30

if it helps draw a right angled triangle and label the 2 speeds then calculate the hypotenuse to find the answer

is that what your asking?

--------------
horizontal velocity = 23cos30 = 19.9
vertical velocity = 23syn30 = 11.5

then it will be wrong =(

 

[tris123]

horizontal speed is constant and not affected by gravity so it stays the same = 19.92

to find speed for vertical 23syn30
THEN
gravity has to be taken into account, so sub that into:
Vy=Uy+AyT

 

[coeyz]

ohh thankyou!!!!!