A person standing on the roof of a building 30 m high drops a ball vertically downward with initial velocity of 500cm/s acceleration due to gravity is 9.8 m/s what will be the velocity of ball after 0.5s ,where will be the ball after 1.5 s , what will be the velocity of the ball while striking the ground .please help me to solve this numerical pleazzzzzzzzzzzzzzzzzz
Answers
Given:-
- Let the downward direction be positive.
U = 500cm/s = 5m/s
S = 30m
A = +9.8m/s^2
1. Using V = U + At, we get
V1 = 5 + (9.8)*(0.5)
V1 = 9.9m/s
2. Using
S = Ut + (1/2)At^2, we get
S1 = 5*(1.5) + (1/2)*(9.8)*(1.5)^2
S1 = 18.525m
3. Using V^2 = U^2 + 2AS, we get
:- V^2 = 5^2 + 2*(9.8)*30
:- V^2 = 613
V = √613
v =24.75 m/s
#The velocity of the ball while striking the ground is 24.75m/s.
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Answer:
Given:-
Let the downward direction be positive.
U = 500cm/s = 5m/s
S = 30m
A = +9.8m/s^2
1. Using V = U + At, we get
V1 = 5 + (9.8)*(0.5)
V1 = 9.9m/s
2. Using
S = Ut + (1/2)At^2, we get
S1 = 5*(1.5) + (1/2)*(9.8)*(1.5)^2
S1 = 18.525m
3. Using V^2 = U^2 + 2AS, we get
:- V^2 = 5^2 + 2*(9.8)*30
:- V^2 = 613
V = √613
v =24.75 m/s
#The velocity of the ball while striking the ground is 24.75m/s.