A hall was dropped from a building and reached the ground in 3s. Show the equati
ID: 1534409 • Letter: A
Question
A hall was dropped from a building and reached the ground in 3s. Show the equations that you use and all calculation to get credit. a) How much was the height of the building? b) How fast will it be going when it hits the ground? c) How much is the acceleration of the ball? Explain A ball is thrown up and it takes 5 seconds to reach maximum height. Show the equation that you use to get credit. a) How fast was it going when I threw it? b) How high up did it go? d) What was the acceleration (magnitude and direction) of the hall going up? Explain. e) What was the acceleration (magnitude and direction) of the ball going down? Explain. f) When was the ball speeding up and slowing down? Explain.Explanation / Answer
1)
(a) t = 3s, let h be the height of the building. since ball was dropped, so u=0, the balls falls under the action of gravity
considering motion in vertical direction
h = ut + (1/2) gt2
h = 0 + (1/2) (9.8) (3)2
h = 44.1 m
(b) Let v be the speed it hits ground with
v = u + gt
v = 0 + (9.8) (3)
v = 29.4 m/s
(c) the ball falls under the action of gravity, so acceleration of ball is
a = g = 9.8 m/s2
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2)
a) Using
v = u+(a*t)
0 = u-(9.8*5)
u = 9.8*5 = 49 m/sec
b) Hmax = u^2/(2g) = 49^2/(2*9.8) = 122.5 m
d) accelaration is 9.8 m/s^2 in the downward direction
explanation :
Since the ball was falling down due to gravitational force only so accelatation is due to gravitational force which is equal to 9.8 m/s^2
e) accelaration is 9.8 m/s^2 in the downward direction
explanation :
Since the ball was falling down due to gravitational force only so accelatation is due to gravitational force which is equal to 9.8 m/s^2
f) while moving up it slows down and while moving down it slows down
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