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$$x=R={u}_{x}T=u\sqrt{\left(\frac{2H}{g}\right)}$$
Problem : A plane flying at the speed of 100 m/s parallel to the ground drops an object from a height of 2 km. Find (i) the time of flight (ii) velocity of the object at the time it strikes the ground and (iii) the horizontal distance traveled by the object.
Solution : The basic approach to solve the problem involves consideration of motion in two mutually perpendicular direction. Here, we consider a coordinate system with the point of release as the origin and down ward direction as the positive y-direction.
(i) Time of flight, T
In vertical direction :
$$\begin{array}{l}{u}_{y}=0\phantom{\rule{2pt}{0ex}},\phantom{\rule{2pt}{0ex}}a=10\phantom{\rule{2pt}{0ex}}m/{s}^{2}\phantom{\rule{2pt}{0ex}},\phantom{\rule{2pt}{0ex}}y=2000\phantom{\rule{2pt}{0ex}}m\phantom{\rule{2pt}{0ex}},\phantom{\rule{2pt}{0ex}}T=?\end{array}$$
Using equation, $y={u}_{y}T+\frac{1}{2}g{T}^{2}$ , we have :
$$\begin{array}{l}\Rightarrow y=\frac{1}{2}g{t}^{2}\\ \Rightarrow T=\sqrt{\left(\frac{2y}{g}\right)}\\ \Rightarrow T=\sqrt{\left(\frac{2x2000}{10}\right)}=20\phantom{\rule{2pt}{0ex}}s\end{array}$$
(ii) Velocity at the ground
We can find the velocity at the time of strike with ground by calculating component velocities at that instant in the two mutually perpendicular directions and finding the resultant (composite) velocity as :
$$\begin{array}{l}v=\sqrt{({{v}_{x}}^{2}+{{v}_{y}}^{2})}\end{array}$$
Initial vertical component of initial velocity (uy) is zero and the object is accelerated down with the acceleration due to gravity. Hence,
$$\begin{array}{l}{v}_{y}={u}_{y}+gt=0+gt=gt\end{array}$$
The component of velocity in the horizontal direction remains unchanged as there is no acceleration in this direction.
$$\begin{array}{l}{v}_{x}={u}_{x}=u\end{array}$$
$$\begin{array}{l}\Rightarrow v=\sqrt{({{v}_{x}}^{2}+{{v}_{y}}^{2})}=\sqrt{({u}^{2}+{g}^{2}{t}^{2})}\end{array}$$
Putting values,
$$\begin{array}{l}\Rightarrow v=\sqrt{({100}^{2}+{10}^{2}x{20}^{2})}=\sqrt{50000}=100\sqrt{5}\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
(iii) Horizontal distance traveled
From consideration of uniform motion in horizontal direction, we have :
$$\begin{array}{l}x={u}_{x}t=ut\end{array}$$
Putting values,
$$\begin{array}{l}\Rightarrow x=R=100x20=2000\phantom{\rule{2pt}{0ex}}m\end{array}$$
A ball is thrown horizontally from a tower at a speed of 40 m/s. The speed of the projectile (in m/s) after 3 seconds, before it touches the ground, is (consider g = 10 $m/{s}^{2}$ ) :
$$\left(a\right)30\phantom{\rule{1em}{0ex}}\left(b\right)40\phantom{\rule{1em}{0ex}}\left(c\right)50\phantom{\rule{1em}{0ex}}\left(d\right)60$$
Here, we consider a reference system whose origin coincides with the point of projection. The downward direction is along y - direction as shown in the figure.
The initial speed of the projectile is equal to the horizontal component of the velocity, which remains unaltered during projectile motion. On the other hand, vertical component of velocity at the start of motion is zero. Thus,
$${u}_{x}=40\phantom{\rule{1em}{0ex}}m/s,\phantom{\rule{1em}{0ex}}{u}_{y}=0$$
Using equation of motion, we have :
$${v}_{y}={u}_{y}+{a}_{y}t$$ $$\Rightarrow {v}_{y}=0+gt=10X3=30\phantom{\rule{1em}{0ex}}m/s$$
Since the horizontal component of velocity remains unaltered, the speed, after 3 second, is :
$$\Rightarrow v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}=\sqrt{{40}^{2}+{30}^{2}}=50\phantom{\rule{1em}{0ex}}m/s$$ Hence, option (c) is correct.
A ball is projected horizontally from a height at a speed of 30 m/s. The time after which the vertical component of velocity becomes equal to horizontal component of velocity is : (consider g = 10 $m/{s}^{2}$ ) :
$$\left(a\right)1s\phantom{\rule{1em}{0ex}}\left(b\right)2s\phantom{\rule{1em}{0ex}}\left(c\right)3s\phantom{\rule{1em}{0ex}}\left(d\right)4s\phantom{\rule{1em}{0ex}}$$
The ball does not have vertical component of velocity when projected. The ball, however, is accelerated downward and gains speed in vertical direction. At certain point of time, the vertical component of velocity equals horizontal component of velocity. At this instant, the angle that the velocity makes with the horizontal is :
$$\mathrm{tan}\theta =\frac{{v}_{y}}{{v}_{x}}=1$$ $$\Rightarrow \theta ={45}^{0}$$
We should note that this particular angle of 45° at any point during the motion, as a matter of fact, signifies that two mutually perpendicular components are equal.
But we know that horizontal component of velocity does not change during the motion. It means that vertical component of velocity at this instant is equal to horizontal component of velocity i.e.
$${v}_{y}={v}_{x}={u}_{x}=30\phantom{\rule{1em}{0ex}}m/s$$
Further, we know that we need to analyze motion in vertical direction to find time as required,
$${v}_{y}={u}_{y}+{a}_{y}t$$ $$\Rightarrow 30=0+10t$$ $$\Rightarrow t=3s$$
Hence, option (c) is correct.
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