Question

A projectile is thrown at an angle of 30° with the horizontal. If the range of the projectile
is 40 m find the initial velocity of projectile​

Answer 1

Answer:

$R = \frac{ {u}^{2} \sin(2 \alpha ) }{g} \\ 40 = \frac{ {u}^{2} sin(2 \times 30)}{10} \\ 400 = {u}^{2} sin60 \\ 800 \div \sqrt{3} = {u}^{2} \\ \\ u = \sqrt{ \frac{800}{ \sqrt{3} } }$

Answer 2

Answer:

$\sqrt{800 \div \sqrt{3} }$

Explanation:

$formula \: for \: horizontal \: range ... \\ r = {u}^{2} \sin(2 \alpha ) \div g \\ we \: know \: the \: value \: of \: \alpha \: and \: r \: are \: 30 \: and \: 40 \: respectvely \\ 40 = {u}^{2} \sin(60) \div 10 \\ we \: know \: \sin(60) = \sqrt{3 \div } 2 \\ (40 \times 10 \times 2) \div \sqrt{3} = {u }^{2} \\ u = \sqrt{800 \div \sqrt{3} }$