Question 1070345
A ball was throw into the air. It path was given by y=2-5tē+28t where y is the height in metres above ground at time t seconds
1)Find the height above ground from with the ball was released.
y(t) = -5t^2 + 28t + 2
y(t) at t = 0
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2)Find the value of t for which the height of the ball was 34 m above ground.
y(t) = -5t^2 + 28t + 2 = 34
Solve for t