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a ball is dropped from the top of the building. the ball takes 05 s to fall past the 3 m length of a window some distance from the top of a building. how fast the ball ball going as it passes the top of the window? how far is the top of the window from the point at which the ball was dropped?
Given : In Case of the ball crossing the window Time = 5s Distance travelled = 3m Acceleration = 10m/s2 Initial velocity = u m/s We have s= ut + 1/2at2 i.e. 3=5u + 1/2*10*25 3= 5u +125 u= -122/5 m/s The ball is moving downwards with speed 24.4 m/s speed as it crosses the top of the window. Now the initial velocity becomes final velocity for the ball before that incident.i.e v = 24.4 m/s & u = 0 m/s for freely falling body. acc. = 10 m/s2 distance travelled is s =v2 /2a = 595.36 / 20 = 59.536 / 2 = 29.768 m This the distance of the window from the top. HOPE YOU HAVE UNDERSTOOD THE PROBLEM
Given : In Case of the ball crossing the window
Time = 5s
Distance travelled = 3m
Acceleration = 10m/s2
Initial velocity = u m/s
We have s= ut + 1/2at2
i.e. 3=5u + 1/2*10*25
3= 5u +125
u= -122/5 m/s
The ball is moving downwards with speed 24.4 m/s speed as it crosses the top of the window.
Now the initial velocity becomes final velocity for the ball before that incident.i.e v = 24.4 m/s & u = 0 m/s for freely falling body.
acc. = 10 m/s2
distance travelled is s =v2 /2a
= 595.36 / 20
= 59.536 / 2
= 29.768 m
This the distance of the window from the top.
HOPE YOU HAVE UNDERSTOOD THE PROBLEM
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