5.4 Projectile motion types  (Page 4/5)

Problem : A projectile is thrown from the top of a building 160 m high, at an angle of 30° with the horizontal at a speed of 40 m/s. Find (i) time of flight (ii) Horizontal distance covered at the end of journey and (iii) the maximum height of the projectile above the ground.

Solution : Unlike horizontal projection, the projectile has a vertical component of initial velocity. This vertical component is acting upwards, which causes the projectile to rise above the point of projection.

Here, we choose the point of projection as the origin and downward direction as the positive y – direction.

(i) Time of flight, T

Here,

$$\begin{array}{l}{u}_y=u\sin \theta =-40\sin {30}^0=-20m/s;\\ y=160m;\\ g=10m/s^2\end{array}$$

Using equation, $y=u_{y}t+\frac{1}{2}g{t}^2$ , we have :

$$\begin{array}{l}\Rightarrow 160=-20t+\frac{1}{2}10t^2\\ \Rightarrow 5t^2-20t-160=0\\ \Rightarrow t^2-4t-32=0\\ \Rightarrow t^2-8t+4t-32=0\\ \Rightarrow t(t-8)+4(t-8)=0\\ \Rightarrow t=-4s\mathrm{or}t=8s\end{array}$$

Neglecting negative value of time, T = 8 s.

(ii) Horizontal distance, R

There is no acceleration in horizontal direction. Using equation for uniform motion,

$$\begin{array}{l}x=u_{x}T\end{array}$$

Here,

$$\begin{array}{l}{u}_x=u\cos \theta =40\cos {30}^0=20\sqrt{3}m/s;\\ T=8s\end{array}$$

$$\begin{array}{l}\Rightarrow x=u_{x}T=20\sqrt{3}x8=160\sqrt{3}m\end{array}$$

(iii) Maximum height, H

The maximum height is the sum of the height of the building ( $H_2$ ) and the height attained by the projectile above the building ( $H_1$ ).

$$\begin{array}{l}H=H_1+H_2\end{array}$$

We consider vertical motion to find the height attained by the projectile above the building ( $H_1$ ).

$$H_1=\frac{u^2{\sin }^2\theta }{2g}$$

$$\Rightarrow H_1=\frac{{\left(40\right)}^2{\sin }^2{30}^0}{2X10}$$

$$\Rightarrow H_1=\frac{{\left(40\right)}^{2}X\left(\frac{1}{4}\right)}{2X10}=20m$$

Thus maximum height, H, is :

$$\begin{array}{l}H=H_1+H_2=20+160=180m\end{array}$$