1.4 Projectile motion types  (Page 5/5)

The time to strike the ground is obtained by considering motion in vertical direction.

Here, $u_y$ = 0 and y=T (total time of flight)

$$y=\frac{1}{2}g{T}^2$$ $$\Rightarrow T=\sqrt{\left(\frac{2y}{g}\right)}$$

Thus, we see that time of flight is independent of both mass and speed of the projectile in the horizontal direction. The two balls, therefore, strike the ground simultaneously.

Hence, option (d) is correct.

We can use the fact that velocities in vertical direction, attained by two bodies through same displacement, are equal. As such, velocity of the second body , on reaching the ground, would also be 3 m/s. Now, there is no acceleration in horizontal direction. The horizontal component of velocity of the second body, therefore, remains constant.

$$v_x=4m/s$$

and

$$v_y=3m/s$$

The resultant velocity of two component velocities in mutually perpendicular directions is :

$$\Rightarrow v=\sqrt{v_x^2+v_y^2}=\sqrt{4^2+3^2}=5m/s$$

First three options pertain to the time of flight. The time of flight depends only the vertical displacement and vertical component of projection velocity. The vertical components of projection in both cases are zero. The time of flight (T) in either case is obtained by considering motion in vertical direction as :

$$y=u_{y}T+\frac{1}{2}{a}_y{T}^2$$ $$\Rightarrow H=0+\frac{1}{2}g{T}^2$$ $$\Rightarrow T=\sqrt{\left(\frac{2H}{g}\right)}$$

Hence, both projectiles reach the ground simultaneously. On the other hand, component of velocity in vertical direction, on reaching the ground is :

$$v_y=u_y+a_y$$ $$\Rightarrow T=0+gT$$

Putting value of T, we have :

$$\Rightarrow v_y=\sqrt{2gH}$$

Projectile (A) has no component of velocity in horizontal direction, whereas projectile (B) has finite component of velocity in horizontal direction. As such, velocities of projectiles and hence the speeds of the particles on reaching the ground are different.

Hence, options (c) and (d) are correct.

The time of flight of a projectile solely depends on vertical component of velocity.

The projectile "A" is thrown up from an elevated point with vertical component of velocity 10 m/s. It travels to the maximum height ( $H_1$ ) and the elevation from the ground( $H_2$ ).

The projectile "B" is thrown down from an elevated point with vertical component of velocity 20 m/s. It travels only the elevation from the ground ( $H_2$ ).

The projectile "C" is thrown down from an elevated point with vertical component of velocity 10 m/s. It travels only the elevation from the ground ( $H_2$ ).

The projectile "D" is thrown up from an elevated point with vertical component of velocity 15 m/s. It travels to the maximum height ( $H_1$ ) and the elevation from the ground( $H_2$ ).

Clearly, projectile "B" and "C" travel the minimum vertical displacement. As vertical downward component of velocity of "B" is greater than that of "C", the projectile "B" takes the least time.

The projectiles "A" and "D" are projected up with same vertical components of velocities i.e. 10 m/s and take same time to travel to reach the ground. As the projectiles are projected up, they take more time than projectiles projected down.

Hence, options (b) and (c) are correct.