A car starts from rest and accelerates at $5\, m / s ^{2}$ At $t=4\, s$, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at $t =6\, s$ ?
(Take $\left. g =10\, m / s ^{2}\right)$
- $20\, m / s , 5\, m / s ^{2}$
- $20\, m / s , 0$
- $20\, \sqrt{2}\, m / s , 0$
- $20\, \sqrt{2}\, m / s , 10\, m / s ^{2}$
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to determine the velocity and acceleration of the ball after it is dropped from the car at a given time.
- First, calculate the velocity of the car just before the ball is dropped (at \(t = 4 \, s\)). The car starts from rest with an acceleration of \(5 \, m/s^2\). We use the formula for velocity:
\(v = u + at\), where \(u = 0\) (initial velocity), \(a = 5 \, m/s^2\) (acceleration), and \(t = 4 \, s\).
Substitute the values:
\(v = 0 + 5 \times 4 = 20 \, m/s\)
- When the person drops the ball at \(t = 4 \, s\), the ball has the same horizontal velocity as the car, \(20 \, m/s\). The ball is subjected to gravity, which affects only the vertical component of its motion, not the horizontal.
- At \(t = 6 \, s\) (2 seconds after being dropped), analyze the motion:
The horizontal component of the velocity remains unchanged as there is no horizontal force acting once it is dropped. Therefore, the horizontal velocity \(v_x\) is still \(20 \, m/s\).
The vertical component of velocity after 2 seconds, \(v_y\), can be calculated as follows:
\(v_y = u_y + g \cdot t\)
Since the ball is dropped, initial vertical velocity \(u_y = 0\):
\(v_y = 0 + 10 \times 2 = 20 \, m/s\)
- Now, find the resultant velocity \(v\) using the Pythagorean theorem:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20)^2 + (20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20 \sqrt{2} \, m/s\)
- The acceleration of the ball after being dropped is due to gravity alone, which is \(10 \, m/s^2\) vertically downward.
Thus, at \(t = 6 \, s\), the ball's velocity is \(20 \sqrt{2} \, m/s\) and its acceleration is \(10 \, m/s^2\).
The correct answer is: \(20 \sqrt{2}\, m / s , 10\, m / s ^{2}\)
Approach Solution -2
$V _{ x }= at =4 \times 5=20 \,m / s$
So horizontal velocity $=20 \,m / s$ (remain constant)
Vertical velocity at $t =6 \sec$
i.e. after $2 sec$ of free fall
$V_{y}=g t=20\, m / s$
So net velocity $=\sqrt{20^{2}+20^{2}}=20 \sqrt{2} \,m / s$
and once it starts falling acceleration is only ' $g$ '
i.e., $10 \,m / s ^{2}$
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Concepts Used:
Projectile Motion
Projectile
A projectile is an object set in flight by applying an external force. The projectile may be thrown up vertically or at an angle to the horizontal. It may be dropped from a position of rest. The only force acting on the projectile during its motion along the flight path is the gravitational force and it is in motion due to its own inertia.
Examples of projectile are: a ball hit by a bat, bullet fired from a gun, shell launched from a launcher, bomb dropped from a plane, etc. It must be noted that a rocket or a missile cannot be considered as a projectile as they are propelled by power.
Trajectory path/ projectile motion
When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the centre of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the motion is called projectile motion or trajectory motion.
In a Projectile Motion, there are two simultaneous independent rectilinear motions:
- Along the x-axis: uniform velocity, responsible for the horizontal (forward) motion of the particle.
- Along y-axis: uniform acceleration, responsible for the vertical (downwards) motion of the particle.
Acceleration in the horizontal projectile motion and vertical projectile motion of a particle: When a particle is projected in the air with some speed, the only force acting on it during its time in the air is the acceleration due to gravity (g). This acceleration acts vertically downward. There is no acceleration in the horizontal direction, which means that the velocity of the particle in the horizontal direction remains constant.
The types of Projectile Motion Formula are:
- Horizontal Distance – x = Vx0t
- Horizontal Velocity – Vx = Vx0
- Vertical Distance, y – Vy0t – ½ gt2
- Vertical Velocity, Vy – Vy0 – gt