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# Motion of a particle thrown upwards and air resistance

There are 5 parts to this question and kindly provide explanation on the derivation of answers.
See attached file for full problem description.

#### Solution Preview

SOLUTION
X
i) v = 0

v

mg

u = 20m/s
x=0

In the fig. is shown an object thrown upwards with an initial velocity of u. Only force acting on the object is the force of gravity acting downwards (mg). Hence, the acceleration of the object is equal to mg/m = g downwards [from Newton's second law : F=ma]. As the object is moving upwards, a downwards acceleration means the object is decelerating. Hence, acceleration of the object = -g .

At the highest point, the object becomes stationary before starting to move downwards. Hence, at the highest point v = 0.

Applying the kinematic equation : v2-u2 = 2as and substituting u = 20 m/s, v = 0 and a = -g = - 9.8 m/s2 we get :

0 - 400 = -2 x 9.8 x h [where h = height of the highest point]

Solving we get : h = 20.4 m

When the object starts falling, the acceleration is in the same direction as the motion. Hence a = +g. Initial velocity now has to be taken as zero since at the highest point the velocity is zero. To determine the velocity of the object as it hits the ground, apply again the equation : v2-u2 = 2as we get :

v2- 0 = 2 x 9.8 x 20.4 or v = 20 m/s [the object will hit the ground with the same velocity with which it was projected]

ii)
X
v = 0

v

...

#### Solution Summary

In this problem the motion of a particle thrown upwards is considered neglecting the air resistance and then taking the air resistance into account.

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