An arrow is projected into air. Its time of flight is 5 s and range 200 m. What is the maximum height reached by it? (Take g = 10 m s-2)

Question

An arrow is projected into air. Its time of flight is 5 s and range 200 m. What is the maximum height reached by it? (Take g = 10 m s-2) (A) 31.25 m (B) 24.5 m (C) 18.25 m (D) 46.75 m

Tardigrade Admin · Accepted Answer

Given, 5=(2u sin θ/g) or (u sin θ /2g)=(5/2) Maximum height =(u2 sin2 θ/2g)=(g/2)×((u2 sin2 θ/g2)) (g/2)×((5/2))2=(10/2)×(25/4)=31.25m

Q. An arrow is projected into air. Its time of flight is 5 s and range 200 m. What is the maximum height reached by it? (Take g = 10 m s^-2)

31.25 m

24.5 m

18.25 m

46.75 m

Given, $5 = \frac{2 u s in θ}{g}$ or $\frac{u s in θ}{2 g} = \frac{5}{2}$
Maximum height $= \frac{u ^{2} s i n ^{2} θ}{2 g} = \frac{g}{2} \times (\frac{u ^{2} s i n ^{2} θ}{g ^{2}})$
$\frac{g}{2} \times (\frac{5}{2})^{2} = \frac{10}{2} \times \frac{25}{4} = 31.25$ m

Q. An arrow is projected into air. Its time of flight is 5 s and range 200 m. What is the maximum height reached by it? (Take g = 10 m s-2)