
A person sitting in an open car moving at constant velocity throws a ball vertically upwards in air. If effect of air resistance is neglected, the ball will fall
(A) Exactly in the hands of the person
(B) Outside the car
(C) In the car behind the person
(D) In the car ahead of the person
Answer
243k+ views
Hint: We will find the distance travelled by the car and the ball in time t using the equation of motion, \[{\text{s = ut + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{a}}{{\text{t}}^{\text{2}}}\] . Then, we will compare both the distances and accordingly we will select the correct option.
Complete step by step answer
Let the velocity of the ball be, v and the velocity of the car be u.
Here the velocity of the car will also be the horizontal velocity of the ball.
Now, let’s calculate the distance travelled by the car in time, t. For this, we will use the second equation of motion.
$s = ut + \dfrac{1}{2}a{t^2}$
Since the car is moving with a constant velocity, acceleration will be zero. Therefore, the distance travelled by the car in time t.
$s = ut + \dfrac{1}{2} \times 0 \times {t^2}$
$ \Rightarrow s = ut$
Similarly, considering that the horizontal velocity of the ball is u. the distance travelled by the ball in time, t will also be s = ut.
This means that the ball will cover the same distance as that of the car in time, t. Hence, the ball will land in the hands of the person who threw the ball.
So, option (A) is the correct answer.
Note It is clearly mentioned in the question that the car is moving with a constant velocity which means that no pseudo force will act on the ball when it is thrown upwards in the air. It is possible that one may get confused and try to solve this kind of question by taking into consideration pseudo force.
Complete step by step answer
Let the velocity of the ball be, v and the velocity of the car be u.
Here the velocity of the car will also be the horizontal velocity of the ball.
Now, let’s calculate the distance travelled by the car in time, t. For this, we will use the second equation of motion.
$s = ut + \dfrac{1}{2}a{t^2}$
Since the car is moving with a constant velocity, acceleration will be zero. Therefore, the distance travelled by the car in time t.
$s = ut + \dfrac{1}{2} \times 0 \times {t^2}$
$ \Rightarrow s = ut$
Similarly, considering that the horizontal velocity of the ball is u. the distance travelled by the ball in time, t will also be s = ut.
This means that the ball will cover the same distance as that of the car in time, t. Hence, the ball will land in the hands of the person who threw the ball.
So, option (A) is the correct answer.
Note It is clearly mentioned in the question that the car is moving with a constant velocity which means that no pseudo force will act on the ball when it is thrown upwards in the air. It is possible that one may get confused and try to solve this kind of question by taking into consideration pseudo force.
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