Q7. A projectile is thrown with a velocity of 10 ms⁻¹ at an angle o

1.	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» <U>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.