1)A falling stones takes 0.2 seconds to fall past a window which is 1 m high.From how far above the top of the window was the stone dropped?
2)A ball is thrown vertically upward to a height of 10 m.Determine the time taken to reach this height and the intial speed of the ball?
3)a stone is thrown vertically upward with a speed of 7 ms per second from the top of a cliff which is 70 m above sea level.Compute the time at which it hits the sea and its speed at that time?
PLEASE SHOW ME THE WAY TO SOLVE ONE BY ONE..
2)A ball is thrown vertically upward to a height of 10 m.Determine the time taken to reach this height and the intial speed of the ball?
3)a stone is thrown vertically upward with a speed of 7 ms per second from the top of a cliff which is 70 m above sea level.Compute the time at which it hits the sea and its speed at that time?
PLEASE SHOW ME THE WAY TO SOLVE ONE BY ONE..
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If u is the velocity at the top of the window and v is the velocity at the bottom of the window
s = (u + v) t/2
(u + v) = 2*1 / 0.2 = 10 m/s ------------------------1
(v- u) = gt = 9.8*0.2 = 1.96 m/ s---------------------2
From 1 and 2
2 v = 11.96 m/s and
2 u = 8.04 m/s
u = 4.02 m/s
The height of window is found form
u² = 2as
s = u²/ 2g = 4.02²/(2*9..8) = 0.83 m
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2.
u² = 2as =2*9.8 * 10
v = 14 m/s
t = u /g = 14/9.8 = 1.43 s
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s = u t + 0.5 at²
Taking upward direction as positive
-70 = 7 t – 0.5*9.8 t²
Solving
t = 4.6 s
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s = (u + v) t/2
(u + v) = 2*1 / 0.2 = 10 m/s ------------------------1
(v- u) = gt = 9.8*0.2 = 1.96 m/ s---------------------2
From 1 and 2
2 v = 11.96 m/s and
2 u = 8.04 m/s
u = 4.02 m/s
The height of window is found form
u² = 2as
s = u²/ 2g = 4.02²/(2*9..8) = 0.83 m
========================
2.
u² = 2as =2*9.8 * 10
v = 14 m/s
t = u /g = 14/9.8 = 1.43 s
========================
s = u t + 0.5 at²
Taking upward direction as positive
-70 = 7 t – 0.5*9.8 t²
Solving
t = 4.6 s
========================
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1) let's say that the time it takes the stone to reach to the top of the window is t0 seconds and the time it take for the stone to reach the bottom of the window is t0+t seconds, where t is the .2 second that it takes for the stone to travel from the top of the window to the bottom of the window. The distance the stone covers from its initial height to the bottom of the window is:
(1/2)*a*(t0+t)^2
The distance that the stone covers from its initial height to the top of the window is:
(1/2)*a*t0^2
In both equations, a = g = 9.8 which is the acceleration due to gravity. The difference between the two distances is 1 meter, or to put it algebraically:
1 = (1/2)*a*(t0+t)^2 - (1/2)*a*t0^2 where t=.2 and a = 9.8
Solve the above equation for t0. Recall that t0 is the time it takes for the stone to reach the top of the window fro its initial height. plug this value in for the second equation.
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2) I think you have to use the equation vf^2 = vi^2 +2*a*(xf-xi)
where vf is the final velocity (zero), vi is initial velocity, xf is final height, xi is initial height, a is 9.8.
You could alternatively use conservation of energy I think:
mgh = (1/2)*m*vi^2 (m's cancel)
Anyway get vi using one of the above methods. Once you do that you get:
10 = (vi)*t - (1/2)*a*t^2 and solve for t
3) 70 + 7*t = 0 solve for t (I assume the speed is given in meters/second)
(1/2)*a*(t0+t)^2
The distance that the stone covers from its initial height to the top of the window is:
(1/2)*a*t0^2
In both equations, a = g = 9.8 which is the acceleration due to gravity. The difference between the two distances is 1 meter, or to put it algebraically:
1 = (1/2)*a*(t0+t)^2 - (1/2)*a*t0^2 where t=.2 and a = 9.8
Solve the above equation for t0. Recall that t0 is the time it takes for the stone to reach the top of the window fro its initial height. plug this value in for the second equation.
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2) I think you have to use the equation vf^2 = vi^2 +2*a*(xf-xi)
where vf is the final velocity (zero), vi is initial velocity, xf is final height, xi is initial height, a is 9.8.
You could alternatively use conservation of energy I think:
mgh = (1/2)*m*vi^2 (m's cancel)
Anyway get vi using one of the above methods. Once you do that you get:
10 = (vi)*t - (1/2)*a*t^2 and solve for t
3) 70 + 7*t = 0 solve for t (I assume the speed is given in meters/second)