Question

a projectile is thrown with a velocity 10ms-1 at an angle of 30 degree with horizontal the value of maximum height gained by it is.
a)1m
b)1.25m
c)2m
d)2.5m​

Answer 1

Answer

• Velicity = 10 m/s
• θ = 30°
• Gravitational Acceleration = 9.8 m/s²

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• Here we are gonna simply substitute the given values to get our final answer. In this answer I've rounded up 1.2755102..... as 1.28

$\sf :\implies Maximum \ Height = \dfrac{u^2sin^2\theta}{2g}$

$\sf :\implies Maximum \ Height = \dfrac{10^2 \times sin^2 \ 30^{\circ}}{2\times 9.8}$

$\qquad\quad\small\mathfrak{\dag \ sin \ 30^{\circ} = \dfrac{1}{2}}$

$\sf :\implies Maximum \ Height = \dfrac{100\times (\frac{1}{2})^2}{19.6}$

$\sf :\implies Maximum \ Height = \dfrac{100\times \frac{1}{4}}{19.6}$

$\sf :\implies Maximum \ Height = \dfrac{25}{19.6}$

$\sf :\implies\underline{\boxed{\sf Maximum \ Height = 1.28 \ m}}$

$\sf \displaystyle\therefore\:\underline{\textsf {The Maximum Height will be \textbf{1.28 m}}}$

Know More!

$\begin{gathered}\begin{gathered}\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}\end{gathered}\end{gathered}$