5.3 Features of projectile motion (application)  (Page 35/3)

Problem : The speed of a particle, projected at 60°, is 20 m/s at the time of projection. Find the time interval for projectile to loose half its initial speed.

Solution : Here, we see that final and initial speeds (not velocity) are subject to given condition. We need to use the given condition with appropriate expressions of speeds for two instants.

$$u=\sqrt{\left(u_x^2+u_y^2\right)}$$

$$v=\sqrt{\left(v_x^2+v_y^2\right)}$$

According to question,

$$u=2v$$

$$\Rightarrow u^2=4v^2$$

$$\Rightarrow u_x^2+u_y^2=4\left(v_x^2+v_y^2\right)$$

But, horizontal component of velocity remains same. Hence,

$$\Rightarrow u_x^2+u_y^2=4\left(u_x^2+v_y^2\right)$$

Rearranging for vertical velocity :

$$v_y^2=u_y^2-3u_x^2={\left(20\sin {60}^0\right)}^2-3X{\left(20\cos {60}^0\right)}^2$$

$$v_y^2=20X\frac{3}{4}-3X20X\frac{1}{4}=0$$

The component of velocity in vertical direction becomes zero for the given condition. This means that the projectile has actually reached the maximum height for the given condition. The time to reach maximum height is half of the time of flight :

$$t=\frac{u\sin \theta }{g}=\frac{20X\sin {60}^0}{10}=\sqrt{3}s$$

Problem : A man can throw a ball to a greatest height denoted by "h". Find the greatest horizontal distance that he can throw the ball (consider g = 10 $m/s^2$ ).

Solution : The first part of the question provides the information about the initial speed. We know that projectile achieves greatest height in vertical throw. Let "u" be the initial speed. We can, now, apply equation of motion " $v^2=u^2+2gy$ " for vertical throw. We use this form of equation as we want to relate initial speed with the greatest height.

Here, v = 0; a = -g

$$\begin{array}{l}\Rightarrow 0=u^2-2gh\\ u^2=2gh\end{array}$$

The projectile, on the other hand, attains greatest horizontal distance for the angle of projection, θ = 45°. Accordingly, the greatest horizontal distance is :

$$\begin{array}{l}{R}_{\max }=\frac{u^2\sin 2X\mathrm{45°}}{g}=\frac{u^2\sin \mathrm{90°}}{g}=\frac{u^2}{g}=\frac{2gh}{g}=2h\end{array}$$