Find (a) time of flight , (b) Max.height (c ) Horizontal range of projectile projected with speed (v) making an angle ` theta` with the horizontal direction from ground.

Explanation

Let, a projectile is projected with an angle `theta` with the horizontal with initial velocity u.
R = range of the projectile and H = maximum height of the projectile
`u cos theta` = horizontal component of the velocity
`u sin theta` = vertical component of the velocity
Then the x and y components of the equation of motion are:
`x=(u cos theta)t ........(i) and y=(u sin theta)t-1//2 gt^(2) .....(iii)`
From equation (1) the value of t can be substituted in equation (2) and the path of the particle in projectile motion is given by the equation, `y=(tan theta)x-gx^(2)//2 (u cos theta)^(2)` which is the equation of parabola. So, the path of the particle in projectile motion is parabolic.
Time of flight: Time of flight
Total time to reach the maximum height and return to ground
Since the time to go up is equal to time to come down (t), T = 2t
At the highest point, velocity is zero, therefore, using the component of velocity equation, we get, `v=u+at, or 0=u sin theta-gT//2 or T=2u sin theta//g`
Maximum Height:
Taking the vertical upward motion in consideration, we get,
`y=H, t=u sin theta//g, u_(y)=u sin theta, a_(y)=-g`
Putting these values in the y component of the equation of motion, we get
`H=u sin theta xx (u sin theta//g) -1/2 g(u sin theta//g)6(2) therefore H=u^(2) sin^(2) theta//2g`
Horizontal Range:
Range (R) is the horizontal distance from the launch point to the point at which the particle returns to the launching plane
`R=u cos theta T= u cos theta (2u sin theta//g)=(u^(2) sin 2theta)/(g) rArr R=(u^(2) sin 2theta)/(g)`