Projection (1 Viewer)

[ToO LaZy ^*]

i'm having trouble getting started....

a ball is thrown from a window that is 16m from the ground. if the angle of projection is 60, initial velocity is 5 and g=10, find:

a) the time taken for the ball to land

b) how far the ball will land from the base of the building

 

[George W. Bush]

Horizontally:
x'' = 0
x' = c = 5cos60= 5/2
x = (5/2)t + k, k=0

Vertically:
y'' = -g
y' = 5sin60 - gt = 15/sqrt2 - gt
y = 15t/sqrt2 - 1/2 gt^2 + 16
etc.

a) at landing, y=0, solve for t.
b) sub that t into x.

 

[Xayma]

It will be a parabola that starts at y=16m, umm otherwise, without the derivation

V_y=5sin60
r=ut+1/2at²
-16=5tsin60-5t²
5t²-5tsin60-16=0

Solve the quadratic for t.

b) V_x=5cos60
=2.5ms<up>-1</sup>
Displacement of x=V_xt
so times t from above by 2.5

 

[ToO LaZy ^*]

Vertically:

y = 15t/sqrt2 t - 1/2 gt^2 + 16

[/B]

is that t after sqrt2 meant to be there?..if so, why?

 

[Xayma]

Originally posted by ToO LaZy ^*
is that t after sqrt2 meant to be there?..if so, why?

Yes it is, however it is (15t/sqrt 2)t, so it is 15t²/sqrt2

 

[ToO LaZy ^*]

Originally posted by Xayma
Yes it is, however it is (15t/sqrt 2)t, so it is 15t²/sqrt2

ahh..righty then.

 

[George W. Bush]

Originally posted by ToO LaZy ^*
is that t after sqrt2 meant to be there?..if so, why?

No, it's not.

My bad.

 

[Xayma]

oops, I didnt see the t² there so I assumed that it was it.

 

[ToO LaZy ^*]

ahhhhh...everything makes sense now..