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 the initial velocity? (g=9.8 m/s2 )

Answer 1

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 the initial velocity? (g=9.8 m/s2 )

Required Solution :-

• The direction of the initial velocity is 30° from the horizontal.

Provided that:-

• Maximum vertical height = 10 m
• Magnitude of initial velocity = 28 m/s

To Calculate:-

• Direction of initial velocity

Using Formula :-

• $\bigstar \red{\underline{\boxed{\pmb{\sf\bf\red{h = \dfrac{u^2 sin^2 \theta}{2g}}}}}}$

Calculation :-

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

$\qquad \pmb {\sf \longrightarrow 10 = \dfrac{784 \times sin^2 \theta}{19.6}}$

$\qquad \pmb{\sf \longrightarrow 196 = 784 \times sin^2 \theta}$

$\qquad \pmb{\sf \longrightarrow sin^2 \theta = \dfrac{196}{784}}$

$\qquad \pmb{\sf \longrightarrow sin^2 \theta = 0.25 }$

$\qquad \pmb{\sf\longrightarrow sin \theta = \sqrt{0.25}}$

$\qquad \pmb{\sf\longrightarrow sin \theta = 0.5}$

$\qquad \pmb{\sf\longrightarrow sin \theta = \dfrac{1}{2}}$

$\qquad \longrightarrow {\pink{\underline{\underline{\pmb{\sf{ \theta = 30^{\circ} }}}}}}$

Therefore–

• The direction of initial velocity is 30° from the horizontal.

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