Question

For an angular projection given to a projectile, find :

1. Maximum height
2. Time of flight
3. Horizontal range.

Answer 1

i. Maximum height:

Step 1: Given data

It is the maximum vertical height attained by the object above the point of projection during its flight denoted by $h$. At maximum height, the projectile will only have a vertical component.

From the diagram, the initial speed will be $u_y=u\sin \theta$

The acceleration will be $-g$.

The final velocity will be $0$

Step 2: Find the displacement

Using the third equation of motion,

$$\begin{gathered} v^2=u^2+2as \\ 0={(u\sin \theta )}^2+2(-g)s \\ \Rightarrow s=\frac{u^2{\sin }^2\theta }{2g} \end{gathered}$$

where $u$ is the initial speed, $v$ is the final speed, $a$ is the acceleration, $s$ is the displacement, $g$ is the acceleration due to gravity.

ii. Time of flight:

Step 1: Given data

The final displacement in the y-direction is 0

The acceleration will be $-g$.

Step 2: Find the time

Using the equation of motion,

$$\begin{gathered} s=u_{y}t+\frac{1}{2}a{t}^2 \\ 0=u\sin \theta t+\frac{1}{2}(-g)t^2 \\ u\sin \theta =\frac{1}{2}gt \\ \Rightarrow t=\frac{2u\sin \theta }{g}-----\left(1\right) \end{gathered}$$

iii. Horizontal range:

Step 1: Given

To calculate the range, we will consider only the horizontal component.

$u_x=u\sin \theta$

Step 2: Find the range

Distance = Speed x Time

$$\begin{gathered} R=u\cos \theta \times t \\ =u\cos \theta \times \frac{2u\sin \theta }{g} \\ =\frac{u^2\times 2\sin \theta \cos \theta }{g} \\ =\frac{u^2\sin 2\theta }{g} \end{gathered}$$

Substituting the value of $t$ from (1).

Hence,

1. Maximum height $s=\frac{u^2{\sin }^2\theta }{2g}$
2. Time of flight $t=\frac{2u\sin \theta }{g}$
3. Horizontal range $R=\frac{u^2\sin 2\theta }{g}$