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The height h (in feet) of an object thrown vertically upward (after t seconds) is given...
The height h (in feet) of an object thrown vertically upward (after t seconds) is given by the function h = -16t^2 + vt + s.
A ball is thrown upward with an initial velocity of 112 ft/sec from an initial height of 128 ft, and allowed to fall on the ground.
a) what is the maximum height it reached?
b) At what time did it hit the ground?
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The height h (in feet) of an object thrown vertically upward (after t seconds) is given by the function: `h = -16t^2 + vt + s`
Assuming s to be the initial height, from where the ball is thrown, the function h will be maximum where h' is zero and h" is negative.
h' = (-16)*2t + v = -32t+v (also check h" = -32, which is always negative, hence it canhaveonly one extreme value, i.e. the maximum) Putting the condition for the maximum, -32t+v = 0 or, t = 3.5 s.
So, the ball will reach its maximum height at 3.5 seconds. Height attained at 3.5 seconds is
`h = -16(3.5)^2 + 112*3.5 + 128` = 324 ft (from the point of throwing).
Time to hit the ground can be obtained from the laws of motion,
`d = ut + 1/2 a*t^2`
here, d = (128+324) = 452 ft., u = 0 (the ball comes to a standstill at the peak), a = acceleratio due to gravity we have to find t.
Putting the values, we get `452 = 0*t+1/2 * 32*t^2`
`rArr t^2 = 452/16`
`rArr t = 5.315 ` s
Therefore, the bal will reach the ground at (3.5+5.3)= 8.8 s after being thrown upwards.
Posted by llltkl on June 13, 2013 at 3:14 AM (Answer #1)
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