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Consider a ball thrown vertically up. Taking air resistance into account, would you expect the time during which the ball rises to be longer or shorter than the time during which it falls? Why?

6 years ago

We also assume that the downward vectors are represented by a negative sign while the upward vectors by a positive sign.

The force at any point on the ball during the upward motion is given as:

Where a is the net acceleration of the ball on its vertical motion, m represents the mass of the ball and g the acceleration due to gravity on Earth.

The negative sign in the equation above accounts for the net direction of the acceleration.

Consider the motion of the ball when it is moving downwards under the action of the gravitational force of Earth. Let us assume that the net force on the ball at any instant be given by F while the net force from the air drag be represented by F

However during the downward motion, the force of friction is in opposite direction relative to the force of gravitation . Therefore the net acceleration in this case is:

where represents the mass of the ball, and g is the acceleration due to gravity on Earth.

By comparing the acceleration in the upward and the downward motion one can deduce that the acceleration during the upward journey of the ball is larger in magnitude than the downward journey.

Now, the acceleration of the ball, defined as the rate of change of velocity over certain is larger when the ball is moving upward than when it is moving downwards. Therefore the time for the journey (upward or downward) depends inversely on the magnitude of acceleration.

Thus, for the upward motion the time of flight will be less relatively accounting for the high acceleration whereas in the downward motion the time of flight will be more.

3 years ago

Let the ball be thrown upwards. Let F_{1 }be the force of air resistance. Also let mg be the weight or force acting downwards while the ball is going upwards. THEN,

Net Force F= −(mg+F_{1)}

Let a_{1} be the net acceleration of the ball while going upwards.

= ma_{1}= −mg−F_{1}

=a_{1}= −(g+F_{1}/m) ........... 1

While falling down , let a_{2} be the net acceleration. In this case the air resisitance is acting upwards and motion is downwards with force mg acting downwards.

=ma_{2}=mg−F_{1}

=a_{2}= (mg−F_{1})/m

= a_{2}=g−F_{1}/m ............ 2

FROM Equation 1 and 2;

a_{1} is greater than a_{2}.

ACCELERATION IS INVERSELY PROPORTIONAL TO THE TIME .

Therefore, time of ascend is smaller than time of descend .

Hope it helps you!!!!

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