A man throws a ball vertically upward and it rises thruogh 20m and returns to his hand .what was the initial velocity of the ball and for how much time (T)it remain in air [g=10m/s^2]
Answers
Answered by
59
h = 20m
a = -10m/s^2
v = 0m/s
v^2 - u^2 = 2as
0 - u^2 = 2*-10*20
-u^2 = -400
u^2 = 400
hence, u= 20m/s
total time = 2* time of ascend
v = u + at
0 = 20 + (-10t)
20 = 10t
hence, t = 2s
total time = 2* time of ascend
= 2*2
= 4s
Here you go
a = -10m/s^2
v = 0m/s
v^2 - u^2 = 2as
0 - u^2 = 2*-10*20
-u^2 = -400
u^2 = 400
hence, u= 20m/s
total time = 2* time of ascend
v = u + at
0 = 20 + (-10t)
20 = 10t
hence, t = 2s
total time = 2* time of ascend
= 2*2
= 4s
Here you go
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Answered by
17
The ball is thrown vertically upwards and attains a height of 20m hence it's final velocity is zero. Acceleration will be negative as it is opposite to the direction of motion and equal to acceleration due to gravity.
here S=20, A=-10, V=0
* Calculation of Initial Velocity *
Using the formula V^2=U^2+2AS
0^2 = U^2 + 2×(-10)×20
0=U^2-400
transposing 400 to left side
400 = U^2
20^2 = U^2
Hence U=20 m/s
* Calculation of Time *
Using the formula V=U+AT
0 = 20 + (-10)×T
0 = 20 - 10T
Transposing 10T to left side
10T = 20
T = 2 Sec
here S=20, A=-10, V=0
* Calculation of Initial Velocity *
Using the formula V^2=U^2+2AS
0^2 = U^2 + 2×(-10)×20
0=U^2-400
transposing 400 to left side
400 = U^2
20^2 = U^2
Hence U=20 m/s
* Calculation of Time *
Using the formula V=U+AT
0 = 20 + (-10)×T
0 = 20 - 10T
Transposing 10T to left side
10T = 20
T = 2 Sec
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