Asked by jeff
In a movie a character jumps off a balcony at an angel of 25 degree 10 meters above the ground and land in a window that is 5 meters away and 9 meters above the ground. What is the intial velocity?
Answers
Answered by
Steve
The initial velocity has horizontal and vertical components:
Vx = v cos(25) = .9063V
Vx = v sin(25) = .4226V
The position has horizontal and vertical components:
Px = Vx * t
Py = 10 - 4.9t^2 + Vy*t
How long does it take to travel the 5m across?
5 = .9063V t
t = 5/.9063V
V = 5/.9063t
How long does it take to drop the 1 meter in height?
-1 = 10 - 4.9t^2 + .4226V*t
-1 = 10 - 4.9t^2 + .4226*5/.9063
-1 = 10 - 4.9t^2 + 2.3315
4.9t^2 = 13.3315
t = sqrt(13.3315/4.9) = 1.65s
To cover 5m in 1.65s, Vh = 3.03, so <b>V = 3.34m/s</b>
Check on position:
10 - 4.9*1.65^2 + .4226 * 3.34 * 1.65 = -1.01m
Vx = v cos(25) = .9063V
Vx = v sin(25) = .4226V
The position has horizontal and vertical components:
Px = Vx * t
Py = 10 - 4.9t^2 + Vy*t
How long does it take to travel the 5m across?
5 = .9063V t
t = 5/.9063V
V = 5/.9063t
How long does it take to drop the 1 meter in height?
-1 = 10 - 4.9t^2 + .4226V*t
-1 = 10 - 4.9t^2 + .4226*5/.9063
-1 = 10 - 4.9t^2 + 2.3315
4.9t^2 = 13.3315
t = sqrt(13.3315/4.9) = 1.65s
To cover 5m in 1.65s, Vh = 3.03, so <b>V = 3.34m/s</b>
Check on position:
10 - 4.9*1.65^2 + .4226 * 3.34 * 1.65 = -1.01m
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