Question

a ball is projected with velocity 10 m/sec at angle of 30o with the horizontal surface. the range of the projectile is

Answer 1

The ball’s initial velocity of 10 m/s can be broken down into a horizontal component vx0, and a vertical component, vy0.

vx0=(10cos30°)m/s, or about 8.66 m/s

vy0=(10sin30°) m/s, or 5 m/s

Neglecting air resistance — which we will definitely be there, as addressing it would be well outside the scope of this simple problem — vx is constant, and vy changes due to gravitational acceleration:

vx(t)=vx0=(10cos30°) m/s

vy(t)=vy0+gt, where g=−9.81 m/s2 — negative since +y is positive, thus (5−9.81t) m/s

At t=1 s, vx=(10cos30°) m/s ≈ 8.66 m/s, while vy=(5−9.81(1)) m/s =−4.81 m/s

Combining those components, v=v2x+v2y−−−−−−√≈8.662+(−4.81)2−−−−−−−−−−−−−√ m/s≈9.9 m/s

Answer 2

Angle =w

R=?

U=10

W=30

R= u^2sin2w/g

R=(10)^2sin2(30)/10

R=100×√3/10×2

R=5√3m

Hope this will help you