Quick Projectile Question? (1 Viewer)

Zokunu · Nov 26, 2013

A ball was initally thrown horizontally and landed 15s at an angle of 15 degrees. Find the:

a) Initial Velocity
b) FINAL Horizontal Velocity
c) FINAL Vertical Velocity
d) Displacement.

panda15 · Nov 26, 2013

$i)\text{Vertical Velocity: }v_{y}=u+at$
$\text{u=0, a=9.8 down, t=15}$
$v_{y}=147ms^{-1}$
$tan15=\frac{147}{v_{x}}$
$\text{Since fired horizontally, initial velocity=horizontal velocity, therefore initial velocity=}548.6ms^{-1}$
$ii)\text{Final horizontal velocity = Initial horizontal velocity }=548.6ms^{-1}$
$iii)v_{y}=0-9.8\times15=147ms^{-1}\text{ down}$
$iv)\text{Horizontal displacement: }s=548.6\times15=8229m$
$\text{Vertical displacement: }s=ut+\frac{1}{2}at^{2}$
$\text{u=0, t=15, a=-9.8, so s=}-1102.5}$
$\tex{Displacement: }\sqrt{(8229)^{2}+(-1102.5)^{2}}=8302.5m$

Zokunu · Nov 27, 2013

panda15 said:
$i)\text{Vertical Velocity: }v_{y}=u+at$
$\text{u=0, a=9.8 down, t=15}$
$v_{y}=147ms^{-1}$
$tan15=\frac{147}{v_{x}}$
$\text{Since fired horizontally, initial velocity=horizontal velocity, therefore initial velocity=}548.6ms^{-1}$
$ii)\text{Final horizontal velocity = Initial horizontal velocity }=548.6ms^{-1}$
$iii)v_{y}=0-9.8\times15=147ms^{-1}\text{ down}$
$iv)\text{Horizontal displacement: }s=548.6\times15=8229m$
$\text{Vertical displacement: }s=ut+\frac{1}{2}at^{2}$
$\text{u=0, t=15, a=-9.8, so s=}-1102.5}$
$\tex{Displacement: }\sqrt{(8229)^{2}+(-1102.5)^{2}}=8302.5m$

Cheers man

Zokunu · Dec 9, 2013

Bump...I know this is abit late, but apparently, the answer to the FINAL vertical velocity and displacement is completely wrong. Can someone please help me out?

Fizzy_Cyst · Dec 9, 2013

It doesnt say if angle is 15 degrees to horiz or vertical. Previous working assumed horiz, perhaps it is to vertical, such that Vx = 147Tan15

Zokunu · Dec 9, 2013

Fizzy_Cyst said:
It doesnt say if angle is 15 degrees to horiz or vertical. Previous working assumed horiz, perhaps it is to vertical, such that Vx = 147Tan15

w8 no, i don't think so.... a and b (Vx =548.6) is completely right. The problem is C and D is just....wrong.

panda15 · Dec 9, 2013

Fizzy_Cyst said:
It doesnt say if angle is 15 degrees to horiz or vertical. Previous working assumed horiz, perhaps it is to vertical, such that Vx = 147Tan15

Isn't it safe to assume that the angle specified to the horizontal unless stated otherwise? Only because HSC questions generally say something like "Fired at an angle of x degrees" when referring to the horizontal angle, and has very similar wording to OP's question.

panda15 · Dec 9, 2013

Zokunu said:
w8 no, i don't think so.... a and b (Vx =548.6) is completely right. The problem is C and D is just....wrong.

What do you have as the answers for iii) and iv)? I'm not certain about part iv) because the question is quite vague, but I'm fairly sure my working for iii) is correct.

Zokunu · Dec 10, 2013

panda15 said:
What do you have as the answers for iii) and iv)? I'm not certain about part iv) because the question is quite vague, but I'm fairly sure my working for iii) is correct.

oh SOZ , iii) is correct. Question 4 is completely wrong but....my bad.

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