Problem 1 - projectile motion

A projectile is launched at an angle α from a horizontal level. Prove that the time of rising upwards to the peak is equal to the time of falling down to the staring level.

Solution:

For the reader’s convenience we repeat here Fig. M2.3 form Ch. M2.1

 

 

 

The time of rising to the peak (maximum height H) was found in M2.1 as

t_r = v_0y / g      [1]

or, after taking into account the angle of launching

t_r = v₀ sin α / g      [1a]

 

The peak (maximum height)  is at a distance H  from the starting level given by

       [2]

Now we have to calculate the time of falling of the projectile from this height. We know that motion in y (vertical) direction is independent of motion in any other direction, so we do not have to take into account the simultaneous motion of the projectile in the x direction.

At the peak of its trajectory an object has the vertical component of velocity

v_y = 0,  

so we will calculate the time of the free fall from that distance H.

Displacement in motion with constant acceleration and with initial velocity equal zero is

D = (1/2) a t²   [3]

 

Substituting H for D and gravity acceleration g for a we have

H = (1/2) g t_f²   [4]

We denoted the time as t_f, to emphasize that it is the time of falling from a given height.

Substituting for H the right hand side of Eq. 2 we get the equation

    [5]

Solving Eq. 5 with respect to time t_f, gives

t_f = v₀ sin α / g                 [6]

So the time of rising (Eq. 6) is equal the time of falling (Eq. 1a)

 

t_r = t_f 

 

and this is what we were asked to prove.

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