Grade 12th passMechanics
Kinematics question of univUniver physics plz answer fast
Kinematics question of univUniver physics plz answer fast
1 Answer
Arun
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.
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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