Question

A ball is thrown from a field with an initial speed of 12 $m{s}^{-1}$ at an angle of ${60}^0$ with the horizontal. Write the projectile equation of the projectile.

• A. $y=x2\sqrt{3}-\frac{g{x}^2}{52}$
• B. $y=x\sqrt{7}-\frac{g{x}^2}{72}$
• C. $y=x\sqrt{3}-\frac{g{x}^2}{72}$
• D. $y=x\sqrt{3}-\frac{g{x}^2}{27}$

Answer 1

The correct option is C $y=x\sqrt{3}-\frac{g{x}^2}{72}$

The equation of a projectile motion is a parabola , which is given by:

$y=x\tan \theta -\left(\frac{g}{2u^2{\cos }^2\theta }\right)x^2$

where, $x=$ horizontal distance travelled by object in time $t$

$y=$vertical distance travelled in time $t$

$\theta =$ angle of projection

$u=$ velocity of projection

Given: $u=12m/s,\theta ={60}^o$

Hence, $y=x\tan 60-\frac{g{x}^2}{2\times {12}^2{\cos }^{2}60}$

$y=x\sqrt{3}-\frac{g{x}^2}{2\times 144\times 1/4}$

$y=x\sqrt{3}-\frac{g{x}^2}{72}$