About 18.7736250 seconds assuming that air friction is zero. I used a graphing calculator.. I'm not sure about what the formula for figuring this out is.
The time it takes for a ball to drop from 36 feet can be calculated using the formula for free fall: ( t = \sqrt{\frac{2h}{g}} ), where ( h ) is the height (36 feet) and ( g ) is the acceleration due to gravity (approximately 32 feet per second squared). Plugging in the values, ( t = \sqrt{\frac{2 \times 36}{32}} ), which gives approximately 1.5 seconds. Therefore, it takes about 1.5 seconds for the ball to drop from 36 feet.
It would take 12.75 seconds to fall 2600 feet with gravity, s= 1/2 g t2
45.5 mph
15 miles/hour (5280 feet/1 mile)(1 hour/3600 seconds) = 22 feet per second
Well the average terminal velocity ofr an average sized jumper is 120mph. I fall at 120mph and fall 10,000 feet in 45/50 seconds so I guess about 213 feet per second.
An object in free fall will fall approximately 64 feet in 2 seconds.
The initial velocity of the ball is 16 feet per second when thrown upward. The velocity decreases as the ball travels upward due to gravity until it reaches its peak and starts to fall back down.
0.3270833333333333
1,100 to 1,300 feet.
Ignoring air resistance, that would be about 145 feet.
1.364 seconds to travel 50 feet at 25mph.
0.977 seconds.
5 seconds
A ball thrown at 55mph will take ~.558 seconds to go 45 feet. A ball will take ~.558 seconds to travel 54 feet at a speed of 66mph. So, the 55mph pitch from 45 feet is simulating 66mph from 54 feet.
It would take 12.75 seconds to fall 2600 feet with gravity, s= 1/2 g t2
If the air isn't slowing you down and the only force on you is the force of gravity,then you fall 2,712 feet in 12.98 seconds. (rounded)
For the average skydiver, the first 15 seconds would cover 2,000 feet. So jumping from 7,000 feet would put him at 5,000 feet in 15 seconds.