23. 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/s".
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Answer:
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*ss
=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=35m/s
by: sruthiandsunmega
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