Marks: 4.00 / 1.00 A particle in thrown vertically upwards with speed a m/s reaches a maximum height b m. Acceleration due to gravity is g. Then select incorrect alternative A a 3a speed at time in m/s 4g 4 B. b speed at is 4 vза m/s 2 С 3a 5a At time & particle will be at the 4g4g' same height from the ground. a 3a Velocity of the particle at -& 2g2g same D is
Answers
Answer:
Solution
(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.
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