Question

Q7. 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 *
0/2
(a) 1 m
(b) 1.25 m
(c) 2 m​

Answer 1

Given:

• Initial velocity (u) = 10m/s.

• Angle of projection = 30°.

To Find:

• Maximum height of projectile.

Solution:

We know that,

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

Substituting values,

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

$\implies\bf{ \frac{100 \times 0.5 \times 0.5}{19.6} }$

$\implies\bf{ \frac{100 \times 0.25}{19.6} }$

$\implies\bf{\dfrac{\cancel{25}}{\cancel{19.6}}}$

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

Hence,

• Value of maximum height gained is 1.275m.