Question

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 1

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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