# Expression  for Time of Flight, Horizontal Range and Maximum height

Time of flight: It is the total time for which the projectile remains in air. We know, at the end of time of flight, the projectile reaches the point on the ground. So, vertical height gained by the y = 0.

From vertical equation of motion, we have:

$$y = (usin \theta )T – \frac{1}{2}gT^2$$

Now,

$$0 = (usin \theta )T – \frac{1}{2}gT^2$$

$$\therefore T = \frac{2usin \theta }{g}…….(i)$$

Equation (i) gives the time of flight of the projectile for velocity of projection u at an angle θ.

Horizontal range: The total distance covered by the body in projectile is called horizontal range.

Horizontal range = Horizontal component of velocity × time of flight

$$or, R = ucos \theta \times \frac{2usin \theta }{g}$$

$$or, R = \frac{u^2sin2 \theta }{g}$$

Maximum horizontal range: The horizontal range will be maximum when sin2θ is maximum.

\begin{align*} i.e., \sin 2\theta &= 1 = \sin 90^o \\ \text {or,} \: 2\theta &= 90^o \\ \text {or,} \: \theta &= 45^o \\ \end{align*}

Maximum Height: The vertical component of velocity at maximum height is 0. So,

\begin{align*} V_y^2 &= (u\sin \theta)^2 – 2gH \\ 0 &= u^2\sin ^\theta -2gH \\ H &= \frac {u^2\sin ^2 \theta }{2g} \\ \end{align*}

Two angle of projections for the same range:

The range R for velocity of projection u and angle of projection θ is:

\begin{align*} R &= \frac {u^2\sin 2 \theta }{g} \end{align*}

If (90-θ) be the another angle of projection,  then:

\begin{align}  \\ R_2 &= \frac {u^2 \sin 2(90 – \theta)}{g} =\frac {u^2 \sin (180 – 2\theta )}{g} = \frac {u^2 \sin 2\theta }{g} \\  \therefore R_1 = R_2 \\ \end{align}

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#### 5 Responses to “Expression  for Time of Flight, Horizontal Range and Maximum height”

1. Vamsi

I can able to understand every question in physics. But i am able to understand how can we solve, which formulae should use in that condition .so please help me