SOLUTION: When a stone is projectad into the air with a vertical speed of 25m/s,its height in meters above the ground after t seconds is given by h=26+5t^
Draw the graph of h=26t-5t^ for v
Question 977919: When a stone is projectad into the air with a vertical speed of 25m/s,its height in meters above the ground after t seconds is given by h=26+5t^
Draw the graph of h=26t-5t^ for values of t from 0 to 6 taking 2cm to represent 10units of h. From your graph estimate
1)After how many seconds the stone will hit the ground
the ground
2)the greatest height
3)for how many seconds is the stone more than 30 meters above the ground
4)find the speed of the stone at 3seconds Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! When a stone is projected into the air with a vertical speed of 25m/s, its height in meters above the ground after t seconds is given by h=26+5t^
Draw the graph of h=26t-5t^ for values of t from 0 to 6 taking 2cm to represent 10units of h. From your graph estimate
h=26+5t^ ************** This is wrong. It's -5t^2, not plus.
h=26t-5t^ ************* This would be correct if you add the exponent 2.
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You say 25 m/sec, but the equations say 26 ???
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Using 25 m/sec for the launch: (notice the exponent 2)
1)After how many seconds the stone will hit the ground
the ground
It impacts at h(t) = 0
-5t^2 + 25t = 0
Solve for t.
(t = 0 is the launch)
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2)the greatest height
Max height is the vertex of the parabola, at t = -b/2a
t = -25/-10 = 2.5 seconds after launch
max ht = h(2.5)
3)for how many seconds is the stone more than 30 meters above the ground
-5t^2 + 25t = 30
Solve for t.
height > 30 is the difference between the 2 times, going up, then descending.
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4)find the speed of the stone at 3 seconds
Launch speed = 25 m/sec
Using h(t) = -5t^2 + 25t, acceleration = -10m/sec/sec
25 + 3*(-10) = -5 m/sec (falling at 5 m/sec)
