Question 977919
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:
{{{h(t) = -5t^2 + 25t}}}  (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)