Is this a quadratic Formula if so how do you solve it?
When falling unto the ground from the top of a 75ft building, a ball’s height, S, at time t can be found by
Please show work so I can get an understanding of how to solve
2007-07-25 04:03:38 · 3 answers · asked by DEE H 1 in Science & Mathematics ➔ Mathematics
Your formula didn't show. It should have been S = 75 - 16t². At ground level, S = 0, so solve
0 = 75 - 16t²
16t² = 75
t² = 75/16
t = (5/4)√3 = 2.165 s.
2007-07-25 04:09:52 · answer #1 · answered by Philo 7 · 0⤊ 0⤋
Assume the ball starts with 0 velocity, and ignore air resistance.
The ball's height is then:
S = 75 + (1/2) * g * t^2
g = acceleration of gravity = 32.2 ft/sec^2
To solve for S, just plug in the numbers for g and t.
To solve for t,
t = sqrt( S - 75) * 2 / g
Just plug in the numbers for S and g.
2007-07-25 11:10:43 · answer #2 · answered by morningfoxnorth 6 · 0⤊ 0⤋
You didn't put in the equation!!!
If it can be factored, then factor and solve.
If not, then use the quadratic formula to solve:
x = [ -b +- sqrt(b^2 - 4ac) ] / 2a
where ax^2 + bx + c = 0
2007-07-25 11:07:16 · answer #3 · answered by Becky M 4 · 0⤊ 0⤋