# 1.4 Projectile motion types  (Page 5/5)

 Page 5 / 5

The time of flight is obtained considering motion in vertical direction as :

$y=H={u}_{y}T+\frac{1}{2}g{T}^{2}$

Rearranging, we get a quadratic equation in “T”,

${T}^{2}+\left(\frac{2{u}_{y}}{g}\right)T-\left(\frac{2{H}_{1}}{g}\right)=0$

One of the two values of "T" gives the time of flight in this case. The horizontal range, on the other hand, is given as :

$x=R={u}_{x}T$

## Exercises

Two balls of masses " ${m}_{1}$ ” and “ ${m}_{2}$ ”are thrown from a tower in the horizontal direction at speeds " ${u}_{1}$ ” and “ ${u}_{2}$ ” respectively at the same time. Which of the two balls strikes the ground first?

(a) the ball thrown with greater speed

(b) the ball thrown with lesser speed

(c) the ball with greater mass

(d) the balls strike the ground simultaneously

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)

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

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.

A body dropped from a height "h" strikes the ground with a velocity 3 m/s. Another body of same mass is projected horizontally from the same height with an initial speed of 4 m/s. The final velocity of the second body (in m/s), when it strikes the earth will be :

$\left(a\right)3\phantom{\rule{1em}{0ex}}\left(b\right)4\phantom{\rule{1em}{0ex}}\left(c\right)5\phantom{\rule{1em}{0ex}}\left(d\right)7$

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.

${v}_{x}=4\phantom{\rule{1em}{0ex}}m/s$

and

${v}_{y}=3\phantom{\rule{1em}{0ex}}m/s$

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

$⇒v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}=\sqrt{{4}^{2}+{3}^{2}}=5\phantom{\rule{1em}{0ex}}m/s$

A projectile (A) is dropped from a height and another projectile (B) is projected in horizontal direction with a speed of 5 m/s from the same height. Then the correct statement(s) is(are) :

(a) The projectile (A) reach the ground earlier than projectile (B).

(b) Both projectiles reach the ground with the same speed.

(c) Both projectiles reach the ground simultaneously.

(d) The projectiles reach the ground with different speeds.

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 :

$y={u}_{y}T+\frac{1}{2}{a}_{y}{T}^{2}$ $⇒H=0+\frac{1}{2}g{T}^{2}$ $⇒T=\sqrt{\left(\frac{2H}{g}\right)}$

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

${v}_{y}={u}_{y}+{a}_{y}$ $⇒T=0+gT$

Putting value of T, we have :

$⇒{v}_{y}=\sqrt{2gH}$

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.

Four projectiles, "A","B","C" and "D" are projected from top of a tower with velocities (in m/s) 10 i + 10 j , 10 i - 20 j , -10 i - 10 j and -20 i + 10 j in the coordinate system having point of projection as origin. If "x" and "y" coordinates are in horizontal and vertical directions respectively, then :

(a)Time of flight of C is least.

(b) Time of flight of B is least.

(c) Times of flights of A and D are greatest.

(d) Times of flights of A and D are least.

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.

## Acknowledgment

Author wishes to thank Scott Kravitz, Programming Assistant, Connexions for making suggestion to remove syntax error in the module.

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