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Kinematics question of univUniver physics plz answer fast

Kinematics question of univUniver physics plz answer fast

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Grade:12th pass

1 Answers

Arun
25750 Points
5 years ago
Dear Ankit
 
Find the time taken to fall past the ledge, and since there are no forces acting in the horizontal direction, there will be no horizontal acceration and thus the horizontal component of velocity will be constant, and velocity = distance/time. 

To find the time taken to fall 11.25 m we nust use one of the kinematic equations. What information do we know? 
d = distance = 11.25 m 
a = acceleration = 9.8 m/s^2 
u = initial vertical velocity= 0 
t = time = ? 

So we must use the equation that relates d, u, a and t, or d = ut + 0.5 at^2, and since u is 0, the quation becomes d = 0.5 at^2, which rearranges to t = sqrt(2d/a) = sqrt(2*11.25/9.8) = 1.515 s. 

Horizontal velocity = 1.5 m/1.515 s = 0.99 m/s.

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