Question

A body is projected with a velocity of 50m/s at an angle of 30° with the horizontal .
Find:-
1) time taken by the projectile to attain the maximum height
2)the time of flight of the projectile 3) the maximum height attained by the projectile
4) The range of the projectile​

Answer 1

EXPLANATION.

Body is projected with a velocity of 50 m/s

at an angle of 30° with the horizontal.

$\sf : \implies \: u \: = 50 \: ms {}^{ - 1} \\ \\ \sf : \implies \: angle \: of \: projection \: = 30 \degree$

$\sf \: \orange{{ \underline{2) = time \: of \: flight \: of \: projectile}}} \\ \\ \sf : \implies \: time \: of \: flight \: = t \: = \frac{2u \sin( \theta) }{g} \\ \\ \sf : \implies \: t \: = \frac{2 \times 50 \times \sin( 30 \degree) }{g} \\ \\ \sf : \implies \: t \: = \frac{ \cancel{100} {}^{50} \times \frac{1}{ \cancel{2}} }{g} \\ \\ \sf : \implies \: t \: = 5 \: seconds$

$\sf : \orange{{ \underline{3) = the \: maximum \: height \: attend \: by \: the \: projectile}}} \\ \\ \sf : \implies \: h_{max} = \frac{ {u}^{2} \sin {}^{2} ( \theta) }{2g} \\ \\ \sf : \implies \: h_{max} = \frac{50 \times 50 \times \sin {}^{2} ( 30 \degree) }{2 \times 10} \\ \\ \sf : \implies \: h_{max} = \frac{250 \times \frac{1}{4} }{20} = \frac{250}{80} \\ \\ \sf : \implies \: h_{max} \: = 3.125 \: \: m$

$\sf : \orange{{\underline{4) = the \: range \: of \: the \: projectile}}} \\ \\ \sf : \implies \: R \: = \frac{ {u}^{2} \sin(2 \theta) }{g} \\ \\ \sf : \implies \: R \: = \frac{50 \times 50 \times \sin(2 \times 30 \degree) }{g} \\ \\ \sf : \implies \: R \: = \frac{250 \times \sin(60 \degree) }{g} \\ \\ \sf : \implies \: R \: = \frac{250 \times \sqrt{3} }{10 \times 2} = \frac{250 \times \sqrt{3} }{20} \\ \\ \sf : \implies \: R \: = 12.5 \sqrt{3} \: m$

Answer 2

Answer:

1. Given : u=50ms^-1
2. angle =30°
3. 1ans:____max height_____31.88
4. 2ans___time to "reach"maxheight___2.55
5. 3ans______range_____220.92