Question

a projectile is thrown with the velocity of 10m/s with an angle 30 with horizontal the maximum height ​

Answer 1

Answer :-

Given :-

• u = 10 m/s
• θ = 30°
• g = 9.8 m/s²

To find :-

• Maximum hieght

Solution :-

$\boxed{\rm Maximum \: height = \frac{u^2 sin^2\theta}{2g}}$

Substituting the value in formula -

$\rm Maximum\: height & = \rm\frac{ {10}^{2} {sin}^{2} 30 }{2 \times 9.8} \\ \\ &= \rm\frac{100 \times (\frac{1}{2} )^{2} }{19.6} \\ \\ &= \rm\frac{100 \times \frac{1}{4} }{19.6} \\ \\ &= \rm\frac{25}{19.6} \\ \\ &= 1.25$

$\boxed{\underline{\rm\red{Maximum\: height = 1.25 m} }}$