Question

33. A body is projected with an initial velocity of 20 ms at an angle of 30° with the horizontal. Calculate
(a) maximum height, (b) time taken to reach the maximum height and (c) horizontal range.​

Answer 1

Explanation:

this is the right answer

hop be related to your question

Answer 2

Answer:

(a) maximum height = 5m

(b) time taken to reach the maximum height = 1 sec

(c) horizontal range​= 20$\sqrt{3}$

Explanation:

I think this question is from Projectile motion

Given : Velocity(u) = 20m/s

            angle of projection : 30 degree

             $u_{x} = ucos 30 =10\sqrt{3}$ m/s

             $u_{y}=usin30 = 10 m/s$

a) Maximum Height

$H = (u_{y}) ^{2} / 2g$

$H = 100/20 = 5 m$

b) Time taken to reach maximum height

$t = u_{y}/ g$$t = u_{y}/g$

t = 10/10 = 1 sec

c) Horizonatal Range

$R = 2u_{x}u_{y} / g$

R = $2*10*10\sqrt{3} /10$

R = $20\sqrt{3}$