Question

A projectile is thrown with a velocity of 10 ms⁻¹ at an angle of 30° with horizontal. The value of maximum height gained by it is *​

Answer 1

Answer: 1.275 m

Given that,

• Initial velocity = 10 m/s
• Angle of projection = 30°

To find,

• Maximum height attained by the projectile

Solution:

The formula for calculating the height of a projectile is given as:

$\boxed{\text{Height of a projectile} = \dfrac{u^2 sin^2\theta}{2g}}$

Substituting the given data, we get:

$\implies \text{Height of the projectile} = \dfrac{10^2 (Sin 30)^{2}}{2\times 9.8}\\\\\\\implies \text{Height of the projectile} = \dfrac{ 100 \times 0.5 \times 0.5 }{19.6}\\\\\\\implies \text{Height of the projectile} = \dfrac{25}{19.6} = \boxed{1.275 \:\:m}$

Hence the value of maximum height attained by the projectile is 1.275 m.

Answer 2

Explanation:

Given:

• Initial velocity (u) = 10m/s.

• Angle of projection = 30°.

To Find:

• Maximum height of projectile.

Solution:

We know that,

$\star \: \boxed{\rm\red{Height \: of \: projectile = \frac{ {u}^{2} {sin \theta}^{2} }{2g}}}$

Substituting values,

$:\implies\mathtt{ \frac{( {10)}^{2} \times ( {sin30)}^{2} }{2 \times 9.8} }$

$:\implies\mathtt{ \frac{100 \times 0.5 \times 0.5}{19.6} }$

$:\implies\mathtt{ \frac{100 \times 0.25}{19.6} }$

$:\implies\mathtt{\dfrac{\cancel{25}}{\cancel{19.6}}}$

$\sf\star \: \underbrace\green{Height \: of \: projectile \: = \: 1.275m} \: \star$

Hence, Value of maximum height gained is 1.275m.