Question

a object is projected at 60 to the horizontal with velocity of 20m/s find its position after 0.5 s

Answer 1

It is given to us that angle between projection velocity and the horizontal is 60 degrees.
We also have the projection velocity= 20m/s

We can break the velocity into components along the y and the axis

[tex] v_{x}=20cos60=(20) \frac{1}{2}=10m/s [/tex]

$v_{y}= 20sin60=(20) \frac{\sqrt{3}}{2} = 10 \sqrt{3}$

Time= 0.5 seconds, therefore,
 $s_{x}=(10) \frac{1}{2}= 5$m

For velocity along y-axis, we also have acceleration=g=10m/$s^{2}$
So,
 $s_{y} = 10 \sqrt{3} ( \frac{1}{2})- \frac{1}{2}.10. \frac{1}{2}. \frac{1}{2}$
$s_{y}= 5 \sqrt{3}- \frac{5}{4}$

Now that we have the displacement in components, let us now determine the final displacement,
 s= $\sqrt{s_{x}^{2}+s_{y}^{2}}$
 s=$\sqrt{100+ \frac{25}{16}-\frac{25 \sqrt{3}}{2}}$
 s=$5 \sqrt{4+ \frac{1}{16} -\frac{\sqrt{3}}{2}} = \frac{5}{4} \sqrt{64+1-8 \sqrt{3}}=\frac{5}{4} \sqrt{65-8 \sqrt{3}$