A golf ball is hit at ground level. The ball is observed to reach maximum height above ground level 2s after being hit. 0.88s after reaching this maximum height, the ball is observed to barely clear a fence that is 215ft from where it was hit.
The acceleration of gravity is 32m/s^2.
How high is the fence.
Answer in units of ft
The acceleration of gravity is 32m/s^2.
How high is the fence.
Answer in units of ft
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we can find the initial vertical velocity knowing it takes 2 s to reach max height; the vertical velocity is zero at max height. since gravity reduces the velocity of objects at the rate of 32.2ft/s/s, the object must have had an initial vertical velocity of 2sx32.2ft/s/s = 64.4ft/s
the max height it will achieve is vy^2/2g where vy is the initial vertical velocity; the max height is
64.4^2/2*32.2 = 64.4 ft
now, the ball travels for another 0.88s; in that time, it will fall from its max height by a distance of
dist = 1/2 gt^2 = 1/2*32.2ft/s/s*(0.88s)^2 = 12.5ft
this means the fence is 64.4ft-12.5ft = 51.9ft high
the max height it will achieve is vy^2/2g where vy is the initial vertical velocity; the max height is
64.4^2/2*32.2 = 64.4 ft
now, the ball travels for another 0.88s; in that time, it will fall from its max height by a distance of
dist = 1/2 gt^2 = 1/2*32.2ft/s/s*(0.88s)^2 = 12.5ft
this means the fence is 64.4ft-12.5ft = 51.9ft high
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I question your facts. Acceleration of gravity on this planet is 9.8 m/s^2, or 32 feet/s^2 and since you state the fence is 215ft away, and want an answer in feet, I think it's supposed to be 32 feet/s^2.
Anyways. The distance to the fence is utterly irrelevant. The vertical motion of the ball is independent from its horizontal motion, so it doesn't matter if it goes 215 feet away or 2000 feet away. The only thing that matters is its vertical component, and the time elapsed.
So. Maximum height above ground level h is equal to g*t^2, where t=2s.
h=0.5g*(2s)^2
h = 2g*s^2
h = 2(32 f/s^2)*s^2
h=64 feet.
After reaching maximum height (and zero vertical speed), the ball falls for 0.88s.
d=0.5g*(0.88s)^2
d = (16ft/s^2)(0.88s)^2
d=12.36 feet.
Therefore the fence is 12.36 feet below the maximum height, 64 feet, or 51.6 feet
The fence is 52 feet high (rounding to two significant digits)
Anyways. The distance to the fence is utterly irrelevant. The vertical motion of the ball is independent from its horizontal motion, so it doesn't matter if it goes 215 feet away or 2000 feet away. The only thing that matters is its vertical component, and the time elapsed.
So. Maximum height above ground level h is equal to g*t^2, where t=2s.
h=0.5g*(2s)^2
h = 2g*s^2
h = 2(32 f/s^2)*s^2
h=64 feet.
After reaching maximum height (and zero vertical speed), the ball falls for 0.88s.
d=0.5g*(0.88s)^2
d = (16ft/s^2)(0.88s)^2
d=12.36 feet.
Therefore the fence is 12.36 feet below the maximum height, 64 feet, or 51.6 feet
The fence is 52 feet high (rounding to two significant digits)
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wow man im sorry but i have no clue you must know alot to come up with a question like that and it must take alot of knowledge if thats where you get stumped on a uestion as complicated as that hope that some can answer your question soon