A ball is launched into the sky at 19.6 feet per second from a 58.8 meter tall building. The equation for the ball’s height, h, at time t seconds is h = -4.9t2 + 19.6t + 58.8 . When will the ball strike the ground?
Mathematics · High School · Sun Jan 24 2021
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Given the statement:
A ball is launched into the sky at 19.6 feet per second from a 58.8 meter tall building. The equation for the ball’s height, h, at time t seconds is h = -4.9t^2 + 19.6t + 58.8 .
Determine the time that the ball will strike the ground.
Solution:
In order to determine for the time that the ball will strike the ground, we equate simply take the first derivative of the equation, then solve for t. In order to determine the first derivative, we simply apply the power rule.
h= -4.9t^2 + 19.6t + 58.8
h' = -9.8t + 19.6
Equate to 0, then solve for t.
Transpose -9.8 to the other side of the equation, hence we must take note that in transposing a number, the sign changes.
9.8t = 19.6
Divide both sides by 9.8 in order to determine the value of t.
9.8t/9.8 = 19.6/9.8
t = 2
Final answer:
t = 2 seconds