Question

A ball is kicked at an angle of 30° with the vertical. If the horizontal component of its velocity is 20 meters per second, what is its maximum height and the horizontal range?​

Answer 1

Answer:

if its kicked at 30 with vertical means angle with horizontal is 60.

x component of velocity never changes and is given 20.

do 20=ucostheta

20=ucos60

therefore u = 40 .

now u have u and theta

just put the values and get the ans

..

max height u^2 sin^2theta/2g

range u^2 × 2sintheta cos theta/g

Answer 2

Answer:Hmax = 60m.

Range=80√3 m.

Explanation: The ball makes and angle of 30° with vertical.

Therefore it makes and angle of 60° with the horizontal.

Now, ucos60 = 20 (x component is given)

u = 40m/s

Hmax = (u²sin²60)/2g

= [40² * (3/4)]/2*10

= 60 m.

Range = (u²sin120)/g

= [40² *(√3)/2]/10

= 80√3 m.