Question

A projectile is fired from the top of a cliff of height h above the ocean below. The projectile is fired at an angle$\mathit{\theta }$above the horizontal and with an initial speed${\mathit{v}}_{\mathbf{i}}$. (a) Find a symbolic expression in terms of the variables${\mathit{v}}_{\mathbf{i}}$, g, and u for the time at which the projectile reaches its maximum height. (b) Using the result of part (a), find an expression for the maximum height${\mathit{h}}_{\mathbf{m}\mathbf{a}\mathbf{x}}$above the ocean attained by the projectile in terms of h,${\mathit{v}}_{\mathbf{i}}$,g, and$\mathit{\theta }$.

Short answer

1. Symbolic expression fir the time at which the projectile reaches its maximum height is,$t=\frac{v_i\sin \theta }{g}$ .
2. Expression for the maximum height above the ocean attained by the projectile is, $h_{\max }=h+\frac{{\left(v_i\sin \theta \right)}^2}{2g}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

The height is, h .

The angle of projectile is, $\theta$.

The initial velocity is, $v_i$.

Significance of the projectile fired at an angle

An object's initial velocity can be split into two components when it is launched from rest at an upward angle. These two elements work apart from one another.

Taking the horizontal and vertical components of initial velocity

Take the velocity component in the direction of vertical and horizontal ,

$u_x=v_i\cos \theta$

$u_y=v_i\sin \theta$

And,

Taking the initial coordinates as

$x_i=0$

$y_i=h$

Solving for time t (a)

For the first half of the projectile till it reaches its maximum position,

Hence, final velocity will be 0

$v_y=u_y+a_{y}t$

Substitute all the value in the above equation.

$$\begin{gathered} 0=v_i\sin \theta -gt \\ gt=v_i\sin \theta \\ t=\frac{v_i\sin \theta }{g} \end{gathered}$$

Hence the symbolic expression fir the time at which the projectile reaches its maximum height is, $t=\frac{v_i\sin \theta }{g}$.

Solving for maximum height using this equation of time we just formulated (b)

$h_{\max }=h+u_{y}t+\frac{1}{2}{a}_y{t}^2$

Substitute all the value in the above equation.

$$\begin{gathered} h_{\max }=h+\left(v_i\sin \theta \right)\frac{\left(v_i\sin \theta \right)}{g}-\frac{1}{2}g\frac{{\left(v_i\sin \theta \right)}^2}{g^2} \\ {h}_{\max }=h+\frac{{\left(v_i\sin \theta \right)}^2}{g}\left(1-\frac{1}{2}\right) \\ {h}_{\max }=h+\frac{{\left(v_i\sin \theta \right)}^2}{2g} \end{gathered}$$

Hence the expression for the maximum height above the ocean attained by the projectile is, $h_{\max }=h+\frac{{\left(v_i\sin \theta \right)}^2}{2g}$.