Question

a body is projected with a velocity of 30 m/s at an angle of 60 with horizontal find the mazimum height time of flight and the horizontal range​

Answer 1

Answer

Given :

• Initial velocity $\sf (u) = 30 m/s$
• Angle of projection $\sf (\theta) = 60\degree$
• Acceleration $\sf (g) = 10 m/s^2$

To find :

• Maximum height
• Time of flight
• Horizontal range

Solution :

$\bf\red{Calculating\: \:Maximum \:height (H)}$

$\boxed{\sf{H = \frac{ {u}^{2} {sin}^{2}\theta}{2g}}}$

$\implies$$\sf H = \frac{ {30}^{2} +{sin}^{2}60 \degree}{2g}$

$\implies$$\sf H = \frac{900 + {( \frac{1}{2}) }^{2} }{2 \times 9.8}$

$\implies$$\sf H = \frac{900}{20} \times \frac{3}{5}$

$\implies$$\sf H = \frac{135}{4}$

$\implies$$\sf H = 33.75 m$

$\underline{\underline{\rm\purple{Maximum \: height = 33.75m}}}$

$\bf\red{Calculating \:Time\: of\: flight (T)}$

$\boxed{\sf{T = \frac{2u sin\theta}{g}}}$

$\implies$$\sf T = \frac{2 \times 30 \times sin 60}{10}$

$\implies$$\sf T = 3 \sqrt{3} s$

$\implies$$\sf T = 5.196 s$

$\underline{\underline{\rm\purple{Time\: of \:flight = 5.196 s }}}$

$\bf\red{Calculating \:Horizontal \:Range (R)}$

$\boxed{\sf{R = \frac{ {u}^{2} sin2\theta}{g}}}$

$\implies$$\sf R = \frac{30 \times 30 \times 2 \: sin60 \degree \: cos60\degree}{10}$

$\implies$$\sf R = 45 \sqrt{3}$

$\implies$$\sf R = 77.94m$

$\underline{\underline{\rm\purple{Horizontal \:Range = 77.94 m}}}$