Question

The body is projected initial velocity of (8i+6j)ms^1. The horizhontal range is​

Answer 1

we know that

$also \: from \: (8i + 6j) \\ usin \alpha = 6 \\ ucos \alpha = 8 \\ as \: ucos \alpha = component \: of \: velocity \: in \: x \: direction \\ and \: usin \alpha = component \: of \: velocity \: in \: y \: direction$

there fore

$4h = r \tan( \alpha )$

where h is maximum height attained by projectile

and r is range

to FIND H

$h = u {}^{2} { \sin} {}^{2} ( \alpha ) \div 2g$

we know that 6=usina

$h = 6 {}^{2} \div 2(10)$

h=1.8

$4h = r \tan( \alpha )$

$4(1.8) = r \tan( \alpha )$

$\tan( \alpha ) = \sin( \alpha ) \div \cos( \alpha )$

$\tan( \alpha ) = 6 \div 8$

$7.2 = r \times 6 \div 8 \\ r = 7.2 \times 8 \div 6 \\ range = 9.6m$

hope it helps