Question

to prepare the condition for projectile motion and explain its path maximum height and time of flight​

Answer 1

Answer:

max. height = [v(i)sintheta]^2/ 2g

time of flight =2v(i)sintheta /g

Explanation:

max height=

using  kinematic equation

v(i)t + 1/2 at^2

=

Ymax = [v(i)sin theta] {v(i)sin theta /g } -1/2 g [v(i) sin theta / g]^2

Ymax = v(i)^2sintheta^2 / g  -  v(i)^2sintheta^2 /2g

Ym = [ v(i) sin theta]^2 / 2g

time of flight =

time (minimum) = v(i) sin theta /g

final time= v(i)sintheta /g mulltiplyied by 2

thats it.

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