Question

A ball is projected vertically upward with a speed of 50 m/s. Find (a) the maximum height, (b) the time to reach the maximum height, (c) the speed at half the maximum height. Take g= 10 m/s².Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution :
Initial speed= u=50m/s
g=10m/s2
At maximum height V=0 m/s

Maximum height:
V²-u²=2as
0²-50²=2*-10*s
s=50*50/20
s=125m
∴maximum height reahed =125m
b) time taken to reach maximum height
t=u/g
t=50/10=5 sec
∴Time taken to reach maximum height =5 sec

c) the speed at half the maximum height:
s=125/2=62.5m
v=?
u=50m/s
a=-10m/s²
From  v²-u²=2as
V=√u²+2as
V=√50²+2x(-10)x 62.5
v=√2500-1250
v=√1250=35 m/s

Answer 2

Answer:

a) 125 m

b) 5 s

c)

$25 \sqrt{2}$m/s