Answer:
(a) Initial velocity of ball (u)=50m/s, acceleration of ball =−g,
final velocity at the highest point (v)=0
So applying the 3rd equation of motion we get:
v
2
=u
2
−2gh
max
⇒0=50
2
−2×10×h
max
⇒h
max
=
20
2500
=125m
(b) Let the time required to reach max height be t. Then applying 1st equation of motion we get:
v=u−gt
⇒0=50−10t
⇒t=5s
(c) Let speed at half of max height be V then:
V
2
=50
2
−2g
2
125
⇒V
2
=2500−1250=1250
⇒V=
1250
=35.35m/s.
Step-by-step explanation:
