I have a simple exercise in our Physics lesson relative to our lesson about Free Falling Objects.
The Question is coming in our Exam yet our teacher didn't answer it for us and left us responsible for it. I need to ask here to confirm my solution:
A Ball is thrown vertically upward @ 20 m/s from the top of a 25m building.
1)Find the velocity of the ball @ t(time)= 5
2)What is the velocity of the ball @ Points C and D
3) What is the max height, B
4) What is the total time of the ball in the air.
THE MODEL OF THE SITUATION: http://i41.tinypic.com/ajw136.png
If you need any more info, ask for it, but I think the givens are enough.....I believe you will be using the Kinematic Equations.
The Question is coming in our Exam yet our teacher didn't answer it for us and left us responsible for it. I need to ask here to confirm my solution:
A Ball is thrown vertically upward @ 20 m/s from the top of a 25m building.
1)Find the velocity of the ball @ t(time)= 5
2)What is the velocity of the ball @ Points C and D
3) What is the max height, B
4) What is the total time of the ball in the air.
THE MODEL OF THE SITUATION: http://i41.tinypic.com/ajw136.png
If you need any more info, ask for it, but I think the givens are enough.....I believe you will be using the Kinematic Equations.
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1)
v = v0 - gt
v = 20m/s - (9.8m/s^2)(5s)
v = -29 m/s (negative means downwards)
2)
At point C, vertical displacement=0
v^2 = v0^2 - 2g(Δy)
v = ((20m/s)^2 - 2(9.8m/s^2)(0))^1/2
v = -20 m/s
At point D, vertical displacement=-25m
v^2 = v0^2 - 2g(Δy)
v = ((20m/s)^2 - 2(9.8m/s^2)(-25m))^1/2
v = -29.83 m/s
3)
At point B, v=0
v^2 = v0^2 - 2g(Δy)
0 = 20^2 - 2(9.8)(Δy)
Δy = 20.41 m
4)
vertical displacement=-25m
Δy = v0t - (1/2)g(t^2)
-25 = 20t - (1/2)(9.8)(t^2)
4.9t^2 - 20t - 25 = 0
Using quadratic equation...
t = 5.08 s
v = v0 - gt
v = 20m/s - (9.8m/s^2)(5s)
v = -29 m/s (negative means downwards)
2)
At point C, vertical displacement=0
v^2 = v0^2 - 2g(Δy)
v = ((20m/s)^2 - 2(9.8m/s^2)(0))^1/2
v = -20 m/s
At point D, vertical displacement=-25m
v^2 = v0^2 - 2g(Δy)
v = ((20m/s)^2 - 2(9.8m/s^2)(-25m))^1/2
v = -29.83 m/s
3)
At point B, v=0
v^2 = v0^2 - 2g(Δy)
0 = 20^2 - 2(9.8)(Δy)
Δy = 20.41 m
4)
vertical displacement=-25m
Δy = v0t - (1/2)g(t^2)
-25 = 20t - (1/2)(9.8)(t^2)
4.9t^2 - 20t - 25 = 0
Using quadratic equation...
t = 5.08 s
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Transcript. Free - Falling Objects A basic Physics question . You have two very different objects . A heavy, large object and a light small object . All freely falling objects accelerate with the same acceleration.