Question

A projectile is launched with speed V0 at an angle of theeta 0 above the horizontal. Find an expression

Answer 1

Answer:

We Have Initial Speed  $V_{0}$

Which is Launched at an angle Ф

So, Let's Divide The Component of Speed $V_{0}$ to

• $V_{0}$ cosФ in x-direction
• $V_{0}$ sinФ in y-direction

at Maximum Height The velocity in y-direction Will Be zero

and an Acceleration of g is continuously acting on it

Now Using Kinematics Equation

v = u + at

Here v = 0

u = $V_{0}$ sinФ

a = -g [opposite to the direction of motion ]

So,

t = $V_{0}$ sinФ/g

and Now Total Time To Reach The Ground Will Be T = 2t

So,

[ T = (2$V_{0}$ sinФ)/g ]

Now Using The The Three Kinematics Equation You Can Similarly Find The

Max. Height = ($V_{0}$ sinФ)²/2g

Range = $V_{0}$² sin2Ф/g

Hope This Helped You (´･ω･`)