Question

If 2 objects are shot off with the same initial speed but different angles with respect
to the horizontal axis and undergo projectile motion travelling over flat ground, which
two angles will result in the same range?
(a) 30◦ and 45◦(b) 30◦ and 60◦
(c) 45◦ and 60◦
(d) None of the above

Answer 1

@Thatkid_marc

Answer 2

@thushananth01

Answer 3

@cormacpayne

Answer 4

I think its C

Answer 5

let the initial velocity of those two objects are \(v\)
and let the first object is projected with an angle \[\theta _1\]
and the second object is projected with angle \[\theta_2\]
so if the range of the first object is \(d_1\)
then
\[d_1=\frac{ v^2*\sin(2\theta_1) }{ g }\]
and range of the second object
\[d_2=\frac{ v^2*\sin(2\theta_2) }{ g }\]
now as their range is equal
so
\[d_1=d_2\]
\[\frac{ v^2*\sin(2\theta_1) }{ g }=\frac{ v^2*\sin(2\theta_2) }{ g }\]
\[\sin(2\theta_1)=\sin(2\theta_2)\]
\[2\sin(\theta_1)\cos(\theta_1)=2\sin(\theta_2)\cos(\theta_2)\]
\[\sin(\theta_1)\cos(\theta_1)=\sin(\theta_2)\cos(\theta_2).........[1]\]

now try option (B) which is 30,60
so here \[\theta_1=30 \]
and
\[\theta_2=60\]
so now put \[\theta_1=30 \] in the left hand side of equation [1]
u'll get
\[\sin(\theta_1)\cos(\theta_1)=\frac{ 1 }{ 2 }*\frac{ \sqrt(3) }{ 2}=\frac{ \sqrt(3) }{ 4}\]

and if u put \[\theta_2=60 \]
in the right hand side of equation [1]
then also u'll get\[\frac{ \sqrt(3) }{ 4}\]
so for angle=30,60 their range is same......
hope this helps @malkassir

Answer 6

http://jsfiddle.net/6wyky/
I wonder if this is rounding error.

Answer 7

Yeah, it is definitely rounding error. lol

Answer 8

where's the error @wio

Answer 9

http://jsfiddle.net/5exTk/