Question

The maximum vertical height of a projectile is 10 m. If the magnitude of the initial velocity is 28 m/s. what is the direction of initial velocity​

Answer 1

Answer:-

$\red{\bigstar}$ Direction of initial velocity $\large\leadsto\boxed{\rm\pink{30^{\circ}}}$

• Given:-

Maximum vertical height = 10 m

Magnitude of initial velocity = 28 m/s

• To Find:-

Direction of initial velocity

• Solution:-

We know,

$\large\underline{\boxed{\bf\green{h = \dfrac{u^2 sin^2 \theta}{2g}}}}$

here,

h = vertical height

u = initial velocity

g = acceleration due to gravity

• Substituting in the Formula:-

→ $\sf 10 = \dfrac{(28)^2 \times sin^2 \theta}{2 \times 9.8}$

→ $\sf 10 = \dfrac{784 \times sin^2 \theta}{19.6}$

→ $\sf 10 \times 19.6 = 784 \times sin^2 \theta$

→ $\sf 196 = 784 \times sin^2 \theta$

→ $\sf sin^2 \theta = \dfrac{196}{784}$

→ $\sf sin^2 \theta = 0.25$

→ $\sf sin \theta = \sqrt{0.25}$

→ $\sf sin \theta = 0.5$

→ $\sf sin \theta = \dfrac{1}{2} \dashrightarrow\bf\red{\bigg(sin30^{\circ} = \dfrac{1}{2}\bigg)}$

→ $\large{\bf\blue{\theta = 30^{\circ}}}$

Therefore, the direction of initial velocity is 30° from the horizontal.