Class 11 NEET Physics Motion in a Straight Line Questions

Let total height =H
Time of ascent =T
So, H=uT−

1

2

gT2
Distance covered by ball in time (T−t) sec.
y=u(T−t)−

1

2

g(T−t)2
So distance covered by ball in last t sec.,
h=H−y =[uT−

1

2

gT2]− [u(T−t)−

1

2

g(T−t)2]
By solving and putting T=

u

g

we will get
h=

1

2

gt2