Question

find the Velocity with which a body should be projected at the angel of 60° with the horizontal so that the
maximum height atained is 30m​

Answer 1

Answer:

28.28 or √(800)

Explanation:

max height reached = (u^2 sin^2x)/2g

[x stands for theta or the angle with which the body was projected]

so,

30=(u^2 (sin^2 (60)))/2(10)

30= (u^2 (√3/2)^2)/20

30×20= u^2 × 3/4

600×4/3=u^2

200 ×4 = u^2

800=u^2

√(800)=u

28.28...=u

There you go.

I hope it helps.

:-)