On Mars, if you hit a baseball, the height of the ball at time t would be modeled by the quadratic function h(t)=-1.85t2+20t+1, where t is seconds and h(t) is in metres.
A) When will the ball hit the ground?
B) How long will the ball be above 17m?
For A I got 5.41. But I'm sure that is wrong.
+128475 OK....let's see
A) -1.85t^2 + 20t + 1 = 0
Using the quadratic formula
-20 ±√[ 20^2 - 4 (-1.85) (1) ] -20 ±√ [ 400 + 7.4]
t = _______________________ = _______________
2 (-1.85) -3.7
t = -20 ±√ [407.4]
____________ = 10.86 sec or -.049 sec (reject)
-3.7
Here's a graph [ I used x instead of t ] : https://www.desmos.com/calculator/znjojz3ud3
+128475 For the second one we want to solve this
-1,85t^2 + 20t + 1 = 17 subtract 17 from both sides
-1.85t^2 + 20t - 16 = 0
t = -20 ±√ [ 20^2 - 4(-1.85)(-16) ] -20 ±√ [ 281.6]
__________________________ = ____________ =
-3.7 -3.7
t = .87 sec or t = 9.94 sec
So...it will be above 17 m for [ 9.94 - .87 ] ≈ 9.07 sec
