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height(s)=1/2(u+v)t 200=1/2(0+v)*1
v=400ft/sec




u = initial speed
v=final speed

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If a ball is dropped from a balloon that is 640 feet above the ground and rising at a rate of 48 feet per second find the time the ball is in the air and the speed of the ball when it hits the ground?

Let v be the velocity when the ball is at 640 feets going downwards v = 48 feet /sec let the velocity with which it reaches the ground be u then, u2=v2+2gh g = acc due t ogravity in feet/sq.sec h = 640 feet the time taken to reach the ground = time to return to 640ft + the time to fall from there Time taken to get to the ground is 8 seconds. Final velocity is 208 feet per second downwards


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An engineer on the ground is looking at the top of a building the angle of elevation to the top of the building is 22 degrees the engineer knows the building is 450 ft tall what is the distance?

If the engineer's eye is at ground level, then the distance to the point on the building underneath its highest point is 450/tan(22) ft. If the engineer was standing and his eyes were x ft above the ground, the distance is (450-x)/tan(22) ft.


The height h of a ball thrown into the air with an initial vertical velocity of 48 feet per second from a height of 5 feet above the ground is given by the equation?

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