A ball is projected from top of a tower with the velocity of 5m/s at an angel of 53° to the horizontal. Its speed when it is at a height of 0.45m from the point of projection is?
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Answered by
11
suppose the particle takes t' to move 0.45 m.
Then, the particle has no vertical velocity.
So, 0.45=(1/2)×10×t'=>t'=9/100 s.
Also, v(y)=velocity in the y direction=9/10m/s.
v(x)=constant=5×cos(53)=-5m/s.
hence v=√(25+0.81)=5.08m/s.
Then, the particle has no vertical velocity.
So, 0.45=(1/2)×10×t'=>t'=9/100 s.
Also, v(y)=velocity in the y direction=9/10m/s.
v(x)=constant=5×cos(53)=-5m/s.
hence v=√(25+0.81)=5.08m/s.
imaftab:
Answer is not correct..
correct answer is 4
plssss tell me that if the particle is projected from the building then it will gain a vertical component of velocity
its not correct
Answered by
50
Time taken to reach 0.45 m
T = Distance / Speed
= 0.45 / (uCos53)
= 0.45 / (5 * 3/5)
= 0.15 seconds
Vertical velocity after 4s
v = u + at
v = uSin53 - gt
v = 5*4/5 - 9.8*0.15
v = 2.53 m/s
Resultant velocity when it reaches a height of 0.45 m is
V = sqrt(v_x^2 + v_y^2)
= sqrt(3^2 + 2.53^2)
≈ 4 m/s
T = Distance / Speed
= 0.45 / (uCos53)
= 0.45 / (5 * 3/5)
= 0.15 seconds
Vertical velocity after 4s
v = u + at
v = uSin53 - gt
v = 5*4/5 - 9.8*0.15
v = 2.53 m/s
Resultant velocity when it reaches a height of 0.45 m is
V = sqrt(v_x^2 + v_y^2)
= sqrt(3^2 + 2.53^2)
≈ 4 m/s
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