a stone is dropped from the top of a tower 400 m high find the velocity of the stone when it hit the ground and time taken
Answers
Answer:
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Explanation:
Both the stones move with a uniform acceleration g, the acceleration due to gravity, acting downwards.
Both the stones move with a uniform acceleration g, the acceleration due to gravity, acting downwards. ∴∴ The acceleration of one relative to the other =g-g=0.
Both the stones move with a uniform acceleration g, the acceleration due to gravity, acting downwards. ∴∴ The acceleration of one relative to the other =g-g=0. At time t=0 , the downward velocity of the two stones are, respectively, 0 and −100m/s-100m/s. So, relative velocity =0-(-100)=100 m/s. The stones move with this relative velocity , which is uniform in absence of any relative acceleration.
Both the stones move with a uniform acceleration g, the acceleration due to gravity, acting downwards. ∴∴ The acceleration of one relative to the other =g-g=0. At time t=0 , the downward velocity of the two stones are, respectively, 0 and −100m/s-100m/s. So, relative velocity =0-(-100)=100 m/s. The stones move with this relative velocity , which is uniform in absence of any relative acceleration. The initial distance between the stones=400m.
Both the stones move with a uniform acceleration g, the acceleration due to gravity, acting downwards. ∴∴ The acceleration of one relative to the other =g-g=0. At time t=0 , the downward velocity of the two stones are, respectively, 0 and −100m/s-100m/s. So, relative velocity =0-(-100)=100 m/s. The stones move with this relative velocity , which is uniform in absence of any relative acceleration. The initial distance between the stones=400m. Thus, they will meet after a time t=400100=4s.400100=4s.
Both the stones move with a uniform acceleration g, the acceleration due to gravity, acting downwards. ∴∴ The acceleration of one relative to the other =g-g=0. At time t=0 , the downward velocity of the two stones are, respectively, 0 and −100m/s-100m/s. So, relative velocity =0-(-100)=100 m/s. The stones move with this relative velocity , which is uniform in absence of any relative acceleration. The initial distance between the stones=400m. Thus, they will meet after a time t=400100=4s.400100=4s. The height through which the first stone falls in time t=4 s is,
Both the stones move with a uniform acceleration g, the acceleration due to gravity, acting downwards. ∴∴ The acceleration of one relative to the other =g-g=0. At time t=0 , the downward velocity of the two stones are, respectively, 0 and −100m/s-100m/s. So, relative velocity =0-(-100)=100 m/s. The stones move with this relative velocity , which is uniform in absence of any relative acceleration. The initial distance between the stones=400m. Thus, they will meet after a time t=400100=4s.400100=4s. The height through which the first stone falls in time t=4 s is, h=12gt2=12×9.8×42=78.4mh=12gt2=12×9.8×42=78.4m
Both the stones move with a uniform acceleration g, the acceleration due to gravity, acting downwards. ∴∴ The acceleration of one relative to the other =g-g=0. At time t=0 , the downward velocity of the two stones are, respectively, 0 and −100m/s-100m/s. So, relative velocity =0-(-100)=100 m/s. The stones move with this relative velocity , which is uniform in absence of any relative acceleration. The initial distance between the stones=400m. Thus, they will meet after a time t=400100=4s.400100=4s. The height through which the first stone falls in time t=4 s is, h=12gt2=12×9.8×42=78.4mh=12gt2=12×9.8×42=78.4m ∴∴ The stones meet after 4 s at a height of (400-78.4) , or, 321.6m.