a stone is allowed to fall from the top of a tower 200m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. calculate when and where the two stones will meet
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Answered by
7
distance traveled by the object thrown from the ground.
=> h = 25t-5t^2
distance travelled by the object dropped from the tower, where u = 0
200-h = 5t^2
200-(25t-5t^2) = 5t^2
t = 8 seconds
The bodies meet at the height of:
h = 25t-5t^2
h = 200 - 320 = -120
| h | = 120meters
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=> h = 25t-5t^2
distance travelled by the object dropped from the tower, where u = 0
200-h = 5t^2
200-(25t-5t^2) = 5t^2
t = 8 seconds
The bodies meet at the height of:
h = 25t-5t^2
h = 200 - 320 = -120
| h | = 120meters
GLAD YOU LIKE IT
PLEASE NOTE MY ANSWER AS BRAINLIEST
Answered by
1
Ahoy!
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Height of the tower = 200m
h = 25t - t^2
Distance travelled by dropped object = 200 - h = 5t^2
put the value of h
200 - (25t - t^2) = 5t^2
200 - 25t + t^2 = 5t^2
200 - 25t = 4t^2
t = 8 seconds
put value of t in
h = 25(8) - 5(25)^2
h = - 120m
_______________________________________
____________
Height of the tower = 200m
h = 25t - t^2
Distance travelled by dropped object = 200 - h = 5t^2
put the value of h
200 - (25t - t^2) = 5t^2
200 - 25t + t^2 = 5t^2
200 - 25t = 4t^2
t = 8 seconds
put value of t in
h = 25(8) - 5(25)^2
h = - 120m
_______________________________________
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