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stiv31 [10]
3 years ago
14

A ball is thrown from a height of 43 meters

Mathematics
1 answer:
AveGali [126]3 years ago
4 0
When the ball will hit the ground, the height will be zero. So we need to replace h with 0 in our equation, and solve for t:
0=43-4t-5t^{2}
-5t^{2}-4t+43=0
To solve this equation we are going to use the quadratic formula: t= \frac{-b(+/-) \sqrt{b^{2}-4ac} }{2a}.
From our height equation, we can infer that a=-5, b=-4, and c=43. So lets replace those values in our quadratic formula to find t
t= \frac{-(-4)(+/-) \sqrt{(-4)^{2}-4(-5)(43)} }{2(-5)}
t=- \frac{5}{8} - \frac{ \sqrt{713} }{8} or t=- \frac{5}{8} + \frac{ \sqrt{713} }{8}
t=-3.96 or t=2.71
Since time cannot be negative, t=2.71 is the solution of our equation.

We can conclude that the ball will hit the ground after 2.71 seconds.
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