A ball is dropped from the top of building .The ball takes 0.5 sec to fall past the 3 m length of a window some distance from the top of building ,How fast was the ball going as it passes the top of the window ? How far is top of the window from the point at which ball was dropped ? ( take g = 9.8 m/s2 )
Answers
Answered by
23
Use the formula for distance, knowing the time.
s = (v0 * t) + (0.5 * g * t^2)
m = (v0 * 0.5) + ( 0.5 * 9.8 * 0.5^2)
3 = (v0 *0.5) + (4.9 * 0.25)
3 = (v0 * 0.5) + (1.225)
v0 = (3 - 1.225) / 0.5
v0 = 3.55 m/sec
This is the velocity when the ball reaches the top of the window.
The formula for final velocity of a falling ball is
vf = v0 + (g * t)
3.55 = 0 + (9.8 * t)
t = 3.55 / 9.8
t = 0.362244 seconds
This is the time it took the ball to reach the top of the window (when velocity = 3.55 m/sec0
Determine how far a ball will fall in 0.362244 seconds
s = (v0 * t) + (.5 * g * t^2)
s = 0 + (4.9 * 0.362244^2)
s = 4.9 * 0.12322
s = 0.603778 meters
This is your answer. 0.603778 meters
s = (v0 * t) + (0.5 * g * t^2)
m = (v0 * 0.5) + ( 0.5 * 9.8 * 0.5^2)
3 = (v0 *0.5) + (4.9 * 0.25)
3 = (v0 * 0.5) + (1.225)
v0 = (3 - 1.225) / 0.5
v0 = 3.55 m/sec
This is the velocity when the ball reaches the top of the window.
The formula for final velocity of a falling ball is
vf = v0 + (g * t)
3.55 = 0 + (9.8 * t)
t = 3.55 / 9.8
t = 0.362244 seconds
This is the time it took the ball to reach the top of the window (when velocity = 3.55 m/sec0
Determine how far a ball will fall in 0.362244 seconds
s = (v0 * t) + (.5 * g * t^2)
s = 0 + (4.9 * 0.362244^2)
s = 4.9 * 0.12322
s = 0.603778 meters
This is your answer. 0.603778 meters
Answered by
16
Answer:Vt+Vb= 12m/s
Explanation: V= u+at
Vt-Vb=at=(9.8)(0.5)= 4.9
(V)^2 =(U)^2+2as
(Vb)^2 - (Vt)^2 = 2(9.8)(3)
Now solve it with the formula of
a^2-b^2 = (a+b)(a-b) and put the value of Vb-Vt = 4.9 and simplify and you get the exect result.
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