Question

a body is projected upwards with velocity v 30 m / s at an angle 30 degree with horizontal determine the time of flight ? ( g = 10 m s ^2)​

Answer 1

Answer:

• The time of flight of a body is 5.196 second.

Explanation:

Given that,

• Velocity (u) = 30 m/s
• Acceleration due to gravity (g) = 10 m/s²
• Angle of projection (θ) = 90° - 30° = 60°

As we know that,

Time of flight, $\: \red \bigstar \boxed{\sf T = \frac{2u \: sin\theta}{g}}$

[ Putting values ]

$\sf \leadsto \: T = \frac{2 \times 30 \times \frac{ \sqrt{3} }{2} }{10} \\ \\ \sf \leadsto \: T = \frac{60 \times \frac{ \sqrt{3} }{2} }{10} \\ \\ \sf \leadsto \: T = \frac{6 \times \frac{ \sqrt{3} }{2} }{1} \\ \\ \sf \leadsto \: T = 3 \sqrt{3} \\ \\ \sf \leadsto \: T = 3 \times 1.732 \\ \\ \sf \leadsto \: T = 5.196 \: sec \: \: \green \bigstar$

Answer 2

Explanation:

Given:-

Velocity(u )=30m/s

Acceleration due to gravity =g=$10m/s {}^{2}$

Angle of projection =${\theta}=90-60=30°$

To find:-

Time of flight

Solution:-

as we know that

${:}\longrightarrow$${\boxed {T={\frac {2uSin {\theta}}{g}}}}$

[by putting values]

${:}\longrightarrow$$T={\frac {2×30×{\frac {{\sqrt{3}}}{2}}}{10}}$

${:}\longrightarrow$$T={\frac{{\cancel {60}}×{\frac {{\sqrt {3}}}{2}}}{{\cancel{10}}}}$

${:}\longrightarrow$$T={\frac {{\cancel{6}}×{\frac {{\sqrt {3}}}{{\cancel{2}}}}}{1}}$

${:}\longrightarrow$$T=3 {\sqrt {3}}$

${:}\longrightarrow$$T=3×1.732$

${:}\longrightarrow$${\underline{\boxed{\bf {5.196sec}}}}$