Question

A stone is projected at an angle of 30° with the vertical. If the horizontal component of its velocity is 9.8 m/s, calculate the maximum height attained by the stone and its horizontal range.​

Answer 1

Angle from horizontal

90-30=60°

Given v cos 60=9.8

$9.8 tan(60) = v_{vertical}\\9.8\sqrt{3}=v_{vertical}\\ Max.height=v_{vertical}^2/2g\\=(9.8\sqrt{3})^2/(2*9.8)\\=3*9.8/2\\=3*4.9\\=14.7m\\Thus,max.height=14.7 m\\\\Mark.as.brainliest.$

Answer 2

Answer:

Explanation: α=60° with horizontal i.e. 90-30°

vx=9.8 m/s

But we know that vx=ux=ucosα

So,

u cosα=9.8

u cos(60°)=9.8

u=19.6 ---- (1)

H=u²sin²α/2g

19.6×19.6×sin²60°/2×10

H=14.4m

R= u²2sinαcosα/g

19.6×19.6×2sin60°cos60°/10

R=19.208√3 m