At the edge of a tall building roof, a prankster throws a water balloon straight upward, with his arm outstretched so when it comes down it won't hit him or the building, but will fall to the street below). If he throws it upward vertically with a speed of 8.00 m/s and the building is 95.0 m tall then :
a. How long after he throws it will it hit the street below?
b. What is its speed as it hits?
c. What was it's maximum height above the street below?
9. When the geyser "Old Faithful erupts, you can time from when the water starts to emerge, to when it reaches its maximum height at 3.5 sec. How high is the water going to go? What was the water's original velocity upward at ground level?
a. How long after he throws it will it hit the street below?
b. What is its speed as it hits?
c. What was it's maximum height above the street below?
9. When the geyser "Old Faithful erupts, you can time from when the water starts to emerge, to when it reaches its maximum height at 3.5 sec. How high is the water going to go? What was the water's original velocity upward at ground level?
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a) Distance balloon rises initially = (v^2/2g), = 3.265 metres.
Time it rises = sqrt. (2h/g), = 0.82 secs.
Height it drops from = (95 + 3.265) = 98.265 metres.
Time to drop from max. height = sqrt. (2h/g), = 4.48 secs.
b) Velocity at ground = sqrt. (2gh), = 43.89m/sec.
c) 98.265 metres. (above).
2. Height = 1/2 (t^2 x g), = 60 metres.
Velocity emerging = (gt), = 9.8 x 3.5, = 34.3m/sec.
Time it rises = sqrt. (2h/g), = 0.82 secs.
Height it drops from = (95 + 3.265) = 98.265 metres.
Time to drop from max. height = sqrt. (2h/g), = 4.48 secs.
b) Velocity at ground = sqrt. (2gh), = 43.89m/sec.
c) 98.265 metres. (above).
2. Height = 1/2 (t^2 x g), = 60 metres.
Velocity emerging = (gt), = 9.8 x 3.5, = 34.3m/sec.