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When a body is projected vertically upwards, it decelerates uniformly due to gravity until its velocity reduces to zero at maximum height. After attaining the maximum height, the body then falls back with an increasing velocity. The body must be given an initial velocity and attains a final velocity of zero at its maximum height. Note that the sign of ‘g’ is negative for a vertical projection. This is because the body moves against gravity.
Hence the three equations of linear motion become:
v=u-gt, (from v=u+at)
h=ut-½gt2, (from s=ut+½gt2)
v2 =u2-2gh, (from v2 =u2 -2as)
But at maximum height hmax, v=0. Thus, the three equations reduce to:
i. gt=u,
ii. h=ut-½gt2
iii. u2 =2gh.
From equation (i), the time taken to attain the maximum height is given by;
t=u/g.
Similarly, the initial velocity u and the maximum height attained by the body hmax can be expressed as:
u=gt=(2ghmax)½
And hmax=ut-½gt2 =u2/2g.
When the body finally falls back to its point of projection, the displacement of the body will be zero. Substituting this in equation (ii), we obtain;
0=ut-½gt2
Therefore, 0=t(2u-gt)
And t=0 or t=2u/g, where t=0 is the time at the start of the projection and,
t is this is the total time of flight i.e for both upward projection and fall back. Note that the total time of flight is twice the time taken to attain maximum height.
also, the velocity of the body just before hitting its point of projection as it falls back is the same in magnitude but in opposite direction to its initial velocity; v=-u.
sharon kalunda answered the question on April 24, 2019 at 05:53
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