A ball is thrown vertically upward with a velocity of 20 m/s from top of the building.the height of the point from where the ball is thrown is 25 m from ground.how high will the ball rise ? how long will it be before the ball hits the ground?
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a) At heighest poing velocity will be 0
So by formula v^2 = u^2 - 2gh
(here - is because you are moving against gravity)
So h = 20*20/2*10 = 20 m
So ball will rise up to 25+20 m from the ground.
b) In second part we have to calculate the time of journey from top of 25 m high building to ground.
For this we should use
-h = ut - gt^2/2
[in this case we are using upward direction as + ve and downward as - ve]
So -25 = 20t - 10t^2/2
or 5t^2 - 20t - 25 = 0
or t^2- 4t - 5 = 0
or (t+1)(t-5) = 0
or t = 5 , -1
So t = 5 sec.
So by formula v^2 = u^2 - 2gh
(here - is because you are moving against gravity)
So h = 20*20/2*10 = 20 m
So ball will rise up to 25+20 m from the ground.
b) In second part we have to calculate the time of journey from top of 25 m high building to ground.
For this we should use
-h = ut - gt^2/2
[in this case we are using upward direction as + ve and downward as - ve]
So -25 = 20t - 10t^2/2
or 5t^2 - 20t - 25 = 0
or t^2- 4t - 5 = 0
or (t+1)(t-5) = 0
or t = 5 , -1
So t = 5 sec.
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