Question

(xi) A projectile is thrown at an angle of 30o to the horizontal. What should be the range of initial velocity (u) so that its range will be between 40m and 50m? Assume g = 10 m s -2.

Answer 1

Answer:

20($\sqrt{2}$ / $\sqrt{3}$) < u < 10 ($\sqrt{10}$ / $\sqrt{3}$  )

Explanation:

Range = u^2 sin2A /g

A = 3o degrees

therefore: u^2 = (R x g)/sin 60

u^2 min : 800/$\sqrt{3}$

u min = 20($\sqrt{2}$ / $\sqrt{3}$)

u max = 10 ($\sqrt{10}$ / $\sqrt{3}$  )

I hope it helps